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A MATHEMATICAL 



SOLUTION BOOK 



CONTAINING 



SYSTEMATIC SOLUTIONS OF MANY OF THE 
MOST DIFFICULT PROBLEMS. 



Taken from the Leading Authors on Arithmetic and Algebra, Many Prob- 
lems and Solutions from Geometry, Trigonometry and Calculus, 
Many Problems and Solutions from the Leading Math- 
ematical Journals of the United States, and 
Many Original Problems and Solutions. 



WITH 



NOTES AND EXPLANATIONS 



BY 

B. F. FINKEL, A. M., M. Sc. 
ii 

Member of the London Mathematical Society, Member of the American 

Mathematical Society, Editor of the American Mathematical 

Monthly, and Professor of Mathematics and 

Physics in Drury College. 



THIRD EDITION- REVISED. 



KIBLER & COMPANY, PUBLISHERS, 

Springfield, Mo. 



COPYRIGHT, 1888, 

BY 

B. F. FINKEL, 

IN THE OFFICE OF THE LIBRARIAN OF 
CONGRESS, 

WASHINGTON, D. C. 



PREFACE. 



This work is the outgrowth of eight years' experience in 
teaching in the Public Schools, during which time I have ob- 
served that a work presenting a systematic treatment of solutions 
of problems would be serviceable to both teachers and pupils. 

It is not intended to serve as a key to any work on mathe- 
matics ; but the object of its appearance is to present, for use in 
the schoolroom, such an accurate and logical method of solving 
problems as will best awaken the latent energies of pupils, and 
teach them to be original investigators in the various branches of 
science. 

It will not be denied by any intelligent educator that the so- 
called "Short Cuts" and "Lightning Methods" are positively in- 
jurious to beginners in mathematics. All the "whys" are cut 
out by these methods and the student robbed of the very object 
for which he is studying mathematics ; viz., the devolpment of 
the reasoning faculty and the power to express his thoughts in 
a forcible and logical manner. By pursuing these methods, 
mathematics is made a mere memory drill and when the memory 
fails, all is lost ; whereas, it should be presented in such a way as 
to develop the memory, the imagination, and the reasoning fac- 
ulty. By following out the method pursued in this book, the 
mind will be strengthened in these three powers, besides a taste 
for neatness and a love of the beautiful will be cultivated. 

Any one who can write out systematic solutions of problems 
can resort to "Short Cuts" at pleasure ; but, on the other hand, 
let a student who has done all his work in mathematics by form- 
ulae, "Short Cuts," and "Lightning Methods" attempt to write 
out a systematic solution one in which the work explains 
itself and he will soon convince one of his inability to express 
his thoughts in a logical manner. These so-called "Short Cuts" 
should not be used at all, in the schoolroom. After pupils and 
students have been drilled on the systematic method of solving 
problems, they will be able to solve more problems by short 
methods than they could by having been instructed in all the 
"Short Cuts" and "Lightning Methods" extant. 

It can not be denied that more time is given to, and more 
time wasted in the study of arithmetic in the public schools than 



2 PREFACE. 

in any other branch of study ; and yet, as a rule, no better results 
are obtained in this branch than in any other. The reason of 
this, to my mind, is apparent. Pupils are allowed to combine 
the numbers in such a way as "to get the answer" and that is all 
that is required. They are not required to tell why they do this, 
or why they do that, but, "did you get the answer?" is the 
question. The art of "ciphering" is thus developed at the ex- 
pense of the reasoning faculty. 

The method of solving problems pursued in this book is 
often called the "Step Method." But we might, with equal pro- 
priety, call any orderly manner of doing any thing, the "Step 
Method." There are only two methods of solving problems a 
right method and a wrong method. That is the right method 
which takes up, in logical order, link by link, the chain of rea- 
soning and arrives at the correct result. Any other method is 
wrong and hurtful when pursued by those who are beginners in 
mathematics. 

One solution, thoroughly analyzed and criticised by a class, 
is worth more than a dozen solutions the difficulties of which are 
seen through a cloud of obscurities. 

This book can be used to a great advantage in the class- 
room the problems at the end of each chapter affording ample 
exercise for supplementary work. 

Many of the Formulae in Mensuration have been obtained by 
the aid of the Calculus, the operation alone being indicated. This 
feature of the work will not detract any from its merits for those 
persons who do not understand the Calculus ; for those who do- 
understand the Calculus it will afford an excellent drill to work 
out all the steps taken in obtaining the formulae. Many of the 
formulae can be obtained by elementary geometry and algebra. 
But the Calculus has been used for the sake of presenting the 
beauty and accuracy of that powerful instrument of mathematics. 

In cases in which the formulae lead to series, as in the case 
of the circumference of the ellipse, the rule is given for a near 
approximation. 

It has been the aim to give a solution of every problem 
presenting anything peculiar, and of those which go the rounds 
of the country. Any which have been omitted will receive space 
in future editions of this work. The limits of this book have 
compelled me to omit much curious and valuable matter in 
Higher Mathematics. 

I have taken some problems and solutions from the School 
Visitor, published by John S. Royer; the Mathematical Maga- 
zine, and the Mathematical Visitor, published by Artemas Mar- 
tin, A. M., Ph. D., LL. D.; and the Mathematical Messenger, 
published by G. H. Harvill, by the kind permission of these 
distinguished gentlemen. 



PREFACE. 3 

It remains to acknowledge my indebtedness to Prof. William 
Hoover, A. M., Ph. D., of the Department of Mathematics and 
Astronomy in the Ohio University at Athens, for critically read- 
ing the manuscript of the part treating on Mensuration. 

Hoping that the work will, in a measure, meet the object 
for which it is written, I respectfully submit it to the use of 
my fellow teachers and co-laborers in the field of mathematics. 

Any correction or suggestion will be thankfully received by 
communicating the same to rne. 

THE AUTHOR. 

1 - 

In bringing out a second edition of this work, I am greatly 
indebted to Dr. G. B. M. Zerr for critically reading the work 
with a view to eliminating all errors. 

THE AUTHOR. 

Drury College, Feb. Jp, 1897. 



PREFACE TO SECOND EDITION. 



PREFACE TO THIRD EDITION. 

The hearty reception accorded this book, as is attested by 
the fact that two editions of 1,200 copies each have already been 
sold, encouraged me to bring out this third edition. 

In doing so, I have availed myself of the opportunity of 
making some important corrections, and such changes and im- 
provements as experience and the suggestions of teachers using 
the book have dictated. The very favorable comments on the 
work by some of the most eminent mathematicians in this 
country confirm the opinion that the book is a safe one to 
put into the hands of teachers and students. 

While mathematics is the exact science, yet not every book 
that is written upon it treats of it as though it were such. In- 
deed, until quite recently, there were very few books on Arith- 
metic, Algebra, Geometry or Calculus that were not mere copies 
of the works written a century ago, and in this way the method, 
the spirit, the errors and the solecisms of the past two hundred 
years were preserved and handed down to the present genera- 
tion. At the present time the writers on these subjects are 
breaking away from the beaten paths of tradition, and the re- 
sult, though not wholly apparent, is a healthier and more vig- 
orous mathematical philosophy. Within the last twenty-five 



4 PREFACE. 

years there has set in, in America, a reaction against the spirit 
and the method of previous generations, so that C. A. Laisant, 
in his La Mathematique Phi to sop hie Enseignement : , Paris, 1898, 
says, "No country has made greater progress in mathematics 
during the past twenty-five years than the United States. The 
most of the text-books on Arithmetic, Algebra, Geometry, and 
the Calculus, written within the last five years, are evidence of 
this progress. 

The reaction spoken of was brought about, to some ex- 
tent, by the introduction into our- higher institutions of learn- 
ing of courses of study in mathematics bearing on the wonder- 
ful researches of Abel, Cauchy, Galois, Riemann, Weirstrass, 
and others. This reaction, it may be said, started as early as 
1832, the time when Benjamin Peirce, the first American worthy 
to be ranked with Legendre, Wallis, Abel and the Bernouillis, 
became professor of mathematics and natural philosophy at 
Harvard University. Since that time the mathematical courses 
in our leading Universities have been enlarged and strengthened 
until now the opportunity for research work in mathematics 
as offered, for example, at the University of Chicago, Harvard, 
Yale, Cornell, Johns Hopkins, Princeton, Columbia and others, 
is as good as is to be found anywhere in the world. For ex- 
ample, the following are the subjects offered at Harvard for the 
Academic year 1899-1900: Logarithms, Plane and Spherical 
Trigonometry; Plane Analytical Geometry; Plane and Solid 
Analytical Geometry ; Algebra ; Theory of Equations. Invar- 
iants ; Differential and Integral Calculus ; Modern Methods in 
Geometry. Determinants ; Elements of Mechanics ; Quater- 
nions with application to Geometry and Mechanics ; Theory of 
Curves and Surfaces; Dynamics of a Rigid Body; Trigonomet- 
ric Series. Introduction to Spherical Harmonics. Potential 
Function ; Hydrostatics. Hydrokinematics. Force Functions 
and Velocity-Potential Functions and their uses. Hydroki- 
netics ; Infinite Series and Products ; The Theory of Functions ; 
Albegra. Galois's Theory of Equations; Lie's Theory as ap- 
plied to Differential Equations ; Riemann's Theory of Func- 
tions ; The Calculus of Variations ; Functions Defined by Linear 
Differential Equations ; The Theory of Numbers ; The Theory 
of Planetary Motions; Theory of Surfaces; Linear Associative 
Algebra; the Algebra of Logic; the Plasticity of the Earth; 
Elasticity; and the Elliptic and the Abelian Transcendants. 

While the great activity and real progress in mathematics 
is going on in our higher institutions of learning, a like degree 
of activity is not yet being manifested in many of our colleges 
and academies and the Public Schools in general. It is not 
desirable that the quantity of mathematics studied in our Public 
Schools be increased, but it is desirable that the quality of 



PREFACE. 5 

the teaching should be greatly improved. To bring about this 
result is the aim of this book. 

It does not follow, as is too often supposed, that any one 
familiar with the multiplication table, and able, perhaps, to 
solve a few problems, is quite competent to teach Arithmetic, 
or "Mathematics," as arithmetic is popularly called. The very 
first principles of the subject are of the utmost importance, 
and unless the correct and refined! notions of these principles 
are presented at the first, quite as much time is lost by the 
student in unlearning and freeing himself from erroneous con- 
ceptions as was required in acquiring them. Moreover, no ad- 
vance in those higher modern developments in Mathematics is 
possible by any one having false notions of its first principles. 

As a branch for mental discipline, mathematics, when 
properly taught, has no superior. Other subjects there are 
that are equally beneficial, but none superior. The idea en- 
tertained by many teachers, generally those who have pre- 
pared themselves to teach other subjects, but teach mathematics 
until an opportunity to teach in their special line presents itself 
to them, that mathematics has only commercial value and 
only so much of it should be studied as is needed by the student 
in his business in after life, is pedagogically and psychologically 
wrong. Mathematics has not only commercial value, but edu- 
cational and ethical value as well, and that to a degree not 
excelled by any other science. No other science offers such 
rich opportunity for original investigation and discovery. So far 
from being a perfected and complete body of doctrine "handed 
down from heaven" and incapable of growth, as many sup- 
pose, it is a subject which is being developed at such a mar- 
velous rate that it is impossible for any but the best to keep 
in sight of its ever-increasing and receding boundary. Because, 
therefore, of the great importance of mathematics as an agent 
in disciplining and developing the mind, in advancing the ma- 
terial comforts of man by its application in every department 
of art and invention, in improving ethical ideas, and in culti- 
vating a love for the good; the beautiful, and the true, the 
teachers of mathematics should have the best training possible. 
If this book contributes to the end, that a more comprehensive 
view be taken of mathematics, better services rendered in pre- 
senting its first principles, and greater interest taken in its 
study, I shall be amply rewarded for my labor in its prepa- 
ration. 

In this edition I have added a chapter on Longitude and 
Time, the biographies of a few more mathematicians, several 
hundred more problems for solution, an introduction to the 
study of Geometry, and an introduction to the study of Algebra. 

The list of biographies could have been extended indefi- 
nitely, but the student who becomes interested in the lives of 



6 PREFACE. 

a class of men who have contributed much to the advancement 
of civilization, will find a short sketch of the mathematicians 
from the earliest times down to the present day in Cajori's 
History of Mathematics or Ball's A Short History of Mathematics. 

The biographies which have been added were taken from 
the American Mathematical Mnothly. I have received much aid 
in my remarks on Geometry from Study and Difficulties of 
Mathematics, by Augustus De Morgan. 

It yet remains for me to express my thanks to my colleague 
and friend, Prof. F. A. Hall, of the Department of Greek, for 
making corrections in the Greek terms used in this edition, 

THE AUTHOR. 
Drury College, July, 1899. 



CONTENTS. 



CHAPTER I. 

DEFINITIONS. 



Mathematics classified 11 | 

CHAPTER II. 
NUMERATION AND NOTATION. 



PAGE 

Definitions 11-14 



Numeration defined 14 

French Method defined 14 

English Method defined 14 

Periods of Notation 15 



Notation defined 

Arabic Notation defined 



15 
15 



Roman Notation defined 15 

Ordinal Numbers 15 

Fractions 18 

Irrational Numbers. 20 

Examples 21 



Addition defined 



Subtraction defined 



CHAPTER III. 

ADDITION. 

22 | Examples 23 

CHAPTER IV. 
SUBTRACTION. 

23 | Examples 24 

CHAPTER V. 
MULTIPLICATION. 

Multiplication defined 24 | Examples 25-26 

CHAPTER VI. 
DIVISION. 

Division defined 26 | Examples 27 

CHAPTER VII. 
COMPOUND NUMBERS. 



Definitions 28 

Time Measure 29 

Definitions in Time Measure . 23-31 

Longitude and Time 31-34 

Standard Time 34-35 



The International Date Line . 36-37 

Examples 37-36 

Solutions 38-39 

Examples 39-40 



Divisor defined 41 

Common Divisor defined 41 



Multiple defined 42 

Common Multiple defined 42 



CHAPTER VIII. 
GREATEST COMMON DIVISOR. 

Greatest Com'n Divisor defined. 41 
Examples 41-42 

CHAPTER IX. 
LEAST COMMON MULTIPLE. 

Least Common Multiple defined 42 



Examples 43-44 



8 



CONTENTS. 



CHAPTER X. 
FRACTIONS. 

PAGB 



Definitions 44-46 

Fractions classified 44 



Solutions of Problems. 
Examples 



PAGE 
46-49 
49-52 



CHAPTER XI. 

CIRCULATING DECIMALS. 



54 
55 
55 



IV. 



I. Addition of Circulates 

II. Subtraction of Circulates . . 
III. Multiplication of Circulates 

CHAPTER XII. 
PERCENTAGE. 



Division of Circulates . 56 

Examples 56-57 



.Definitions ....... 57 

Solutions 57-69 

II. Commission . 69 

Definitions ...... .... 69 

Solutions 69-71 

Examples . 72 

Trade Discount 72 

Definitions 72 

Solutions 73-76 

Examples 76 

Profit and Loss 77 



III. 



IV. 



Definitions 77 

Examples 83-84 

V. Stocks and Bonds 84 

Definitions . . . 84 

Solutions 85-95 

Examples 96-97 

VI. Insurance 97 

Definitions 97-98 

Solutions . . 98-101 

Examples 102 



CHAPTER XIII. 

INTEREST. 



II. 



III. 



Simple Interest 103 

Definitions 103 

Solutions 103-106 

True Discount . 106 

Definitions 106 

Solutions 106-107 

Bank Discount 107 

Definitions . . .107 



Solutions 108-109 

IV. Annual Interest 109 

Annual Interest defined 109 

Solutions 110-112 

V, Compound Interest 112 

Compound Int. defined. 112 
Solutions 113-115 



Definitions 
Solutions . 



CHAPTER XIV. 

ANNUITIES. 

. . . . 115 I Examples 
.116-125 



126 



CHAPTER XV. 

MISCELLANEOUS PROBLEMS. 
Solutions . 127-139 



CHAPTER XVI. 
RATIO AND PROPORTION. 

Definitions 139-141 I Problems 

Solutions. ......... 141-144 



... 144-146 



Analysis defined. . . 
Solutions . ... . . 



CHAPTER XVII. 

ANALYSIS. 
. . . . 146 Problems 
146-177 



177-180 



CONTENTS. 



CHAPTER XVIII. 

ALLIGATION. 



PAGE. 

I. Alligation Medial 181 

II. Alligation Alternate. . 181 



Solutions 



PAGE 

181-186 



CHAPTER XIX. 
SYSTEMS OF NOTATION. 



Definitions 187 

Names of Systems 187 



Solutions 188-191 



CHAPTER XX. 
MENSURATION. 



v 
V, 



Definitions 192-197 

Geometrical Magnitudes class- 
ified 192 

I. Parallelogram . . 198-200 

II. Triangles 200-204 

III. Trapezoid 204-205 

IV. Trapezium and Irregular 

Polygons 205 

V. Regular Polygons . . . 205-207 

VI. Circles 207-210 

VII. Rectification of Plane 
Curves and Quadrature of 

Plane Surfaces 210-213 

II. Conic Sections 223 

Definitions 223-224 

1. Ellipse 224-227 

2. Parabola 227-229 

3. Hyperbola 229-232 

IX. Higher Plane Curves ... 233 

1. The Cissoid Diocles. 233-234 

2. The Conchoid of Nicom- 

edes 234-235 

3. The Oval of Cassini 235 

4. The Lemniscate of Ber- 

nouilli 236 

5. The Witch of Agnesi 236-237 

6. The Limacon 237 

7. The Quadratrix 238 

8. The Catenary 238-239 

9. The Tractrix 240 

10. The Syntractrix 240 

II. Roulettes 240 

(a) Cycloids 240-243 

(b) Prolate and Curtate 
Cycloid 243-244 

(c) Epitrochoid and Hy- 
potrochoid 245-248 

X. Plane Spiral 248 

1. Spirals of Archimedes 249 

2. The Reciprocal Spiral 249 

3. The Lituus 250 



4. The Logarithmic Spi- 
ral 250 

XI. Mensuration of Sol- 
ids 251-254 

1. Cylinder 254-255 

2. CylindricUngulas 255-262 

3. Pyramid and Cone 262-266 

4. Conical Ungulas. 266-270 
XII. Sphere 270-276 

XIII. Spheroid 276-278 

1. The Prolate Sphe- 

roid 276-278 

2. The Oblate Sphe- 

roid 278-282 

XIV. Conoids 282 

1. The Parbolic Co- 

noid .... 282-285 

2. Hyperbolic Co- 

noid 285-286 

XV. Quadrature and Cuba- 
ture of Surfaces and 
Solids of Revolution 286 

1. Cycloid 286-287 

2. Cissoid 287-288 

3. Spindles 288-289 

4. Parabolic Spindle 289-290 
XVI. Regular Solids 290 

1. Tetrahedron .... 291-292 

2. Octahedron 292* 

3. Dodecahedron . . . 292-293 

4. Icosahedron 293-294 

XVII. Prismatoid 294-295 

XVIII. Cylindric Rings ... 295-297 
XIX. Miscellaneous Measure- 
ments 297 

1. Masons' and 
Bricklayers' work 297 

2. Gauging 297-298 

3. Lumber Measure . 298 

4. Grain and Hay. . , 298-299 



10 



CONTENTS. 



MENSURATION Concluded. 

PAGE 

XX. Solutions of Miscellaneous Problems 299-345 

Problems 346-354 

Examination Tests : . 354-360 

Problems 361-366 

GEOMETRY. 

Definitions 367 (e) Assumption of the Sphere . . 380 

On Geometric Reasoning 369 (/) of Motion 380 

On the Advantages Derived from On Logic 380 

the Study of Geometry and Laws of Thought 381 

Mathematics in General 370 Law of Converse 383 

Axioms 375 Methods of Reasoning 384 

General Axioms 376 How to Prepare a Lesson in 

Assumptions . 377 Geometry 388 

(a) Assumption of Straight Line 377 Plane Geometry 390 

(b) of the Plane 377 The Three Famous Problems of 

(c) of Parallel Line 377 Antiquity 408 

(d) of the Circle... 379 

ALGEBRA. 

Definitions 415 Arithmetical Fallacies 424 

Solutions of Problems 416-419 Probability 425 

The Quadratic Equation 419 Problems 428 

Indeterminate forms 421 

Biography of Prof. William Hoover 403 

Probability Problems 434-435 

Biography of Dr. Artemas Martin 436 

Biography of Prof. E. B. Seitz 440 

Biography of Rene 7 Descartes 442 

Biography of Leonhard Euler 446 

Biography of Spphus Lie 451 

Biography of Simon Newcomb 454 

Biography of George Bruce Halsted 457 

Biography of Prof. Felix Klein 459 

Biography of Benjamin Peirce 462 

Biography of James Joseph Sylvester s 468 

Biography of Arthur Cayley 475 

Table I 473 

II 478 

478 

J V 479 

V 479 

VI 480 

VII 480 

Example 4gl 



CHAPTER I. 

DEFINITIONS. 

1. Mathematics (p.aftv)p.a.Ttxrj, science) is that science 
which treats of quantity. 



!(1.) Arithmetic. 
(2.) Algebra... 

*.j Geometry.. 



rl. Calculus 

1 2. Quaternions. 

El. Platonic Geometry.. 
2. Analytical Geometry. 
3. Descriptive Geometry. 



Differential. 

Integral. 

Calculus of Variations. 



: a. Pure Geometry. 
b. Conic Sections. i\. Plane Trigon'y. 
c . Trigonometry.. <2. Analytical Trig. 
(3. Spherical " 



'(1.) Mensuration. 

(2.) Surveying. 

(3.) Navigation. 

(4.) Mechanics. 

(5.) Astronomy. 

(6.) Optics. 

(7.) Gunnery. 
^(8.) &c., &c. 

2>. Pure Mathematics treats of magnitude or quantity 
without relation to matter. 

3. Applied Mathematics treats of magnitude as subsist- 
ing in material bodies. 

4. Arithmetic (api^^nx^, from d^etf/io?, a number) is 
the science of numbers and the art of computing by them. 

5. Alyebra (Ar. al, the, and geber, philosopher) is that 
method of mathematical computation in which letters and other 
symbols are employed. 

6. G-eOWietry (yajfj.Tpia, from yU){j.Tpiv to measure 
land, from pa, yf h the earth, and p.Tp~iv, to measure) is the 
science of position and extension. 

7. Calculus ( Calculus, a pebble) is that branch of mathe- 
matics which commands by one general method, the most diffi- 
cult problems of geometry and physics. 



12 FINKEL'S SOLUTION BOOK. 

8. Differential Calculus is that branch of Calculus 
which investigates mathematical questions by measuring the re- 
lation of certain infinitely small quantities called differentials. 

9. Integral Calculus is that branch of Calculus which 
determines the functions from which a given differential has been 
derived. 

1C. Calculus of Variations is that branch of calculus 
in which the laws of dependence which bind the variable quanti- 
ties together are themselves subject to change. 

11. Quaternions (quaternis, from quaterni four each, 
from quator, four) is that branch of algebra which treats of the 
relations of magnitude and position of lines or bodies in space by 
means of the quotient of two direct lines in space, considered as 
depending on a system of four geometrical elements, and as ex- 
pressed by an algebraic symbol of quadrinominal form. 

12. Platonic Geometry is that branch of geometry in 
which the argument is carried forward by a direct inspection of 
the figures themselves, delineated before the eye, or held in the 
imagination. 

13. Pure Geometry is that branch of Platonic geometry 
in which the argument may be practically tested by the aid of 
the compass and the square only. 

14. Conic Sections is that branch of Platonic geometry 
which treats of the curved lines formed by the intersection of a 
cone and a plane. 

15. Trigonometry (rptytovov, triangle, ptrpov, meas- 
ure) is that branch of Platonic geometry which treats of the re- 
lations of the angles and sides of triangles. 

16. Plane Trigonometry is that branch of trigonom- 
etry which treats of the relations of the angles and sides of plane 
triangles. 

17. Analytical Trigonometry is that branch of trig- 
onometry which treats of the general properties and relations of 
trigonometrical functions. 

18. Spherical Trigonometry is that branch of trig- 
onometry which treats of the solution of spherical triangles. 

19. Analytical Geometry is that branch of geometry 
in which the properties and relations of lines and surfaces are in- 
vestigated by the aid of algebraic analysis. 

20. Descriptive Geometry is that branch of geometry 
which seeks the graphic solution of geometrical problems by 
means of projections upon auxiliary planes. 



DEFINITIONS. 13 

21. Mensuration is that branch of applied mathematics 
which treats of the measurment of geometrical'magnitudes. 

22. Surveying is that branch of applied mathematics 
which treats of the art of determining and representing distances, 
areas, and the relative position of points upon the earth's surface. 

23. Navigation is that branch of applied mathematics 
which treats of the art of conducting ships from one place to 
another. 

24:. Mechanics is that branch of applied mathematics 
which treats of the laws of equilibrium and motion. 

25. Astronomy d.ffrpovop.ia^ from affrpov, star and VO/JLOS 
law) is that branch of applied mathematics in which mechan- 
ical principles are used to explain astronomical facts. 

26. Optics (oTtTix-ij, from 8</>ts sight,) is that branch of 
applied mathematics which treats of the laws of light. 

27. GrUnnery is that branch of applied mathematics which 
treats of the theory of projectiles. 

28. A Proposition is a statement of something proposed 
to be done. 

/ rm \ ! Lemma. 

, T^ I, ( a. Ineorem. \ on 

1. Demonstrable. \ ^ p ro ^j em ( 2 - Corollary. 

29. Prop't'n. - 

, , ( a. Axiom. 

2. Indemonstrable. < , ^ 

/ . Postulate. 

30. A Demonstrable Proposition is one that can 
be proved by the aid of reason. 

31. A. Theorem is a truth requiring a proof. 

32. A Lemma is a theorem demonstrated for the purpose 
of using it in the demonstration of another theorem. 

33. A Corollary is a subordinate theorem, the truth of 
which is made evident in the course of the demonstration of a 
more general theorem. 

34. A Problem is a question proposed for solution. 

35. An Indemonstrable Proposition can not be 

proved by any manner of reasoning. 

36. An Axiom is a self-evident truth. 

37. A Postulate is a proposition which states that some- 
thing can be done, and which is so evidently true as to require 
no process of reasoning to show that it is possible to be done. 



14 i-lNKEL'S SOLUTION BOOK. 

38. A Demonstration is the process of reasoning, prov- 
ing the truth of a proposition. 

39. A Solution of a problem is an expressed statement 
showing clearly how the result is obtained. 

40. An Operation is a process of finding, from given 
quantities, others that are known, by simply illustrating the 
solution. 

41. A Rule is a general direction for solving all problems 
of a particular kind. 

42. A Formula is the expression of a general rule or 
principle in algebraic language. 

43. A Scholium is a remark made at the close of a dis- 
cussion, and designed to call attention to some particular feature 
or features of it. 



CHAPTER II. 

NUMERATION AND NOTATION. 

1. Numeration is the art of reading numbers. 

2. There are two methods of numeration ; the French and the 
English. 

3. The French method is that in general use. In this method, 
we begin at the right hand and divide the number into periods 
of three figures each, and give a distinct name to each period. 

4. The English method is that used in Great Britain and the 
British provinces.. In this method, we divide the number (if it 
consists of more than six figures) into periods of six figures each, 
and give a distinct name to each period. The following number 
illustrates the two methods ; the upper division showing how the 
number is read by the English method, and the lower division 
showing how it is read by the French method. 

4th period, 3d period, 2d period, 1st period. 
Trillions. Billions. Millions. Units. 

"^845 678^904 325^47 434^913 

5-3 5 5g -SS ga *.! * p 
*~< ^o? * ^ ^EH ^ 

5. The number expressed in words by the English method, 
'eads thus: 




NUMERATION AND NOTATION. 15 

Eight hundred forty-five trillion, six hundred seventy-eight 
thousand nine hundred four billion, three hundred twenty-five 
thousand one hundred forty-seven million, four hundred thirty- 
four thousand nine hundred thirteen. 

Remark. Use the conjunction and, only in passing over 
the decimal point. It is incorrect to read 456,734 four hundred 
and fifty-six thousand, seven hundred and thirty-four. Omit the 
and's&nd the number will be correctly expressed in words. 

6. The following are the names of the Periods, according to 
the common, or French method: 



First Period, Units. 

Second " Thousands. 

Third " Millions. 

Fourth " Billions. 

Fifth " Trillions, 



Sixth Period, Quadrillions. 
Seventh " Quintillions. 
Eighth " Sextillions. 
Ninth " Septillions. 
Tenth " Octillion. 



Other periods in order are, Nonillions, Decillions, Undecil- 
lions, Duodecillions, Tredecilions, Quatuordecillions Quindecil- 
lions, Sexdecillions, Septendecillions, Octodecillions, Novende- 
cillions, Vigintillions, Primo-Vigintillions, Secundo-vigintillions, 
Tertio-vigintillions, Quarto- vigintillions, Quinto-vigintillions, 
Sexto-vigintillions, Septo-vigintillions, Octo-vigintillions, Nono- 
vigintillions, Trigillions; Primo-Trigillions, Secundo-Trigillions, 
and so on to Quadragillions ; Primo-quadragillions, Secundo- 
quadragillions, and so on to Quinquagillions; Primo-quinqua- 
'gillions, Secundo-quinquagillions, and so on to Sexagillions, 
Pr i mo -sexagil lions, Secundo-sexagillions, and so on to Septua- 
gillions ; Primo-septuagillions, Secundo-septuagillions, and so on 
to Octogillions ; Primo-octogillions, Secundo-octogillions, and 
so on to Nonogillions ; Primo-nonogillions, Secundo-nonogillions, 
and so to Centillions. 

7. Notation is the art of writing numbers. 

There are three methods of expressing numbers ; by words, 
by letters, called the Roman method, and by figures, called the 
Arabic method. 

8. The Roman Notation, so called from its having originated 
with the ancient Romans, uses seven capital letters to express 
numbers; viz., I, V, X, L, C, D, M. 

9. The Arabic Notation, so called from its having been made 
known through the Arabs, uses ten characters to express num- 
bers ; viz., 1, 2, 3, 4, 5 9 6, 7, 8, 9, 0. 

10. Ordinal Numbers. A logical definition of number 
is not easy to give, for the reason that the idea it conveys is a 
simple notion. The clearest idea of what counting and numbers 
mean inay be gained from the observation of children and of 



16 FINKEL'S SOLUTION BOOK. 

nations in the childhood of civilization*. When children count 
or add they use their ringers, or small sticks, or pebbles which 
they adjoin singly to the things to be counted or otherwise to be 
ordinally associated with them. History informs us that the 
Greeks and Romans employed their fingers when they counted 
or added. The reason why the fingers are so universally used 
as a means of numeration is, that everyone possesses a definite 
number, sufficiently large for purposes of computation and that 
they are always at hand. 

Let us consider the row of objects, XXXXXXXX 

XXXXXXXX , with regard to their order, say 

from left to right, freeing our minds from all notions of magni- 
tude. Beginning with any one object in this row, we speak of 
the one we begin with as being the first, the next in order to it 
to the right the second, the next in order to the right of the sec- 
ond the third, and so on. The name or mark we thus attach to 
an object to tell its place in the row is called an integer. This 
process, or operation, of labeling the objects is called counting 
and it is the fundamental operation of mathematics. 
To count objects is to label the objects, not primarily to tell how 
many there aref. In thus labeling the objects, we may replace 
the objects by the fingers, by sticks, by pebbles, by marks, or 
by characters. The method of tallying used at the present time 
is such a method. In counting objects marks are made until 
four are made, then these are crossed with a fifth mark and so 
on. Thus fH-F -ffH fH4. 

Suppose that in counting the objects in the row, we use our 
fingers, and for each object in the row beginning with a certain 
one we bring in correspondence with that object the little fin- 
ger of the right hand, with the next object to the right the next 
to the little finger of the right hand, and so on until an object 
and the thumb of the right hand are brought into correspond- 
ence. For the group of objects thus counted, let us bring into 
correspondence the little finger of the left hand. Now continue 
the counting of the objects of the row as before, and when a 
second group is reached bring into correspondence with this 
group the next to the little finger of the left hand. Continue 
this process until a group of the objects as represented by the 
fingers of the right hand is brought into correspondence with 
the thumb of the left hand. Thus the fingers of the left hand 
represent a group of groups of objects. Bring this group rep- 
resented by the fingers of the left hand into correspondence with 

* Schubert's Mathematical Essays and Recreations. 

f My friend, Dr. William Rullkoetter, told me of a case coming under his personal 
observation, where a farmer, unable to count, but when desirous of knowing: if any of 
his cattle were missing, would have them driven through a gate or past some point 
where he could see them as they passed singly. He would then say, " You are here," 
" and you are here," " and you are here," and so on until all had passed by. In this way 
he was able to tell if any were missing, but not able to tell how many he had. 



NUMERATION AND NOTATION. 17 

the little toe of the right foot. Now continue the process of 
counting the objects and so on as before until the big toe of the 
right foot is brought into correspondence with a group corre- 
sponding to the fingers of the left hand. Thus the toes of the 
right foot represent a group of a group of a group of objects. 
In this manner, we could build up the system of numeration 
called the Quinary, a system in which five objects as represented 
by the fingers of the right hand make a unit or group as repre- 
sented by a finger of the left hand, five groups of five objects as 
represented by the fingers of the left hand make a group as rep- 
resented by the toes of the right foot, and so on. 

The decimal system of numeration may be built up in the 
same way, except that the group of objects corresponding to the 
fingers of both hands would be represented by a toe. After the 
fingers and toes have been exhausted in the process of counting 
the numeration would have to be continued by using small sticks 
or pebbles. It is very probably due to the fact that we have 10 
fingers .that the decimal system was invented. There are, how- 
ever, among the uncivilized nations of the world a number of 
different systems of numeration*. 

At the present time, in labeling objects by the process of 
counting we use the following characters, viz., 1, 2, 3, 4, 5, 
6, 7, 8, 9, etc. 

123456789 labels. 

Thus XXXXXXXXXXXXXXX objects. 

In labeling, we could begin .with the object marked 3 and 
re-label it 1, then re-label 4 as 2 and 5 as 3, and so on. This is 
expressed by writing 

32=1, 42=2, 52=3, 

meaning that if we begin after the object whose old mark was 2, 
then the object which was third becomes first, the object which 
was fourth becomes second, and so on. Beginning after an 
object instead of with it suggests that our original row might 
begin after an object; this object after which the counting begins 
is marked and called the origin. If there are objects to the 
left of the origin, we count them in the same way; except that 
we prefix the sign, , to show that they are to the left, and we 
call the marks so changed negative integers, thus distinguishing 
them from the old marks which we call positive integers. The 
marks are .... 4, 3, 2, --1, 0, 1, 2, 3, 4, 5, 6, 

These marks constitute what is called the natural integer- 
system. 

When an object marked a is to the left of another marked 
a', we say that a comes before a' or is inferior to a', and a' 

*See Conant's Number Concept for a full treatment of the various systems of 
notation. 



18 FINKEL'S SOLUTION BOOK. 

comes after a or is superior to a. These ideas are expressed 
symbolically thus a<a! ', a>a. Here a and a' mean integers, 
positive or negative. 

Objects considered as a succession from left to right are in 
positive order; when considered from right to left, in negative 
order. 

Addition and its inverse operation, Subtraction, are algorithms 
of counting. Multiplication is an algorithm of Addition, and 
Division is an algorithm of Subtraction. Addition, Subtraction, 
Multiplication, and Division are only short methods of counting. 

If we operate on any integer of the natural-integer series by 
any one of the operations of Addition, Subtraction, or Multipli- 
cation, no new integer is produced. With reference to these 
operations the natural integer-series is closed, that is to say, 
there are no breaks in the integer-series into which other inte- 
gers arising from these operations may be inserted. If, how- 
ever, we operate on any one of the integers of the integer-series 
by the operation of Division, the operations in many cases are 
impossible. Suppose we wish to divide 17 by 5. This opera- 
tion is absolutely impossible. *- is a meaningless symbol with 
reference to the fundamental operation of mathematics. But in 
this case, as in the case when negative numbers are introduced 
by the inverse operation, subtraction, we apply a principle called 
by Hankel, "The Principle of the Permanence of Formal Laws," 
and by Schubert, "The Principle of No Exception," viz., That 
every time a newly introduced concept depends upon operations 
previously employed, the propositions holding for these operations 
are assumed to be valid still when they are applied to the new con- 
cepts. In accordance with this principle, we invest the symbol, 
- 1 /-, which has the form of a quotient without its dividend being 
the product of the divisor and any number yet defined, with a 
meaning such that we shall be able to reckon with such apparent 
quotient as with ordinary quotients. This is done by agreeing 
always to put the product of such a quotient form with its 
divisor equal to its dividend. Thus, (^-)Xb=Vj. We thus 
reach the definition of fractions. The concept of fractions may 
also be established as in the next article. 

11. Fractions. Let us now again assume the row of 

0123456789 

objects, XXXXXXXXXXXX. . . . attending to only 

zero, the object from which we begin, and the objects on the 
right of it. Suppose we re-label the alternate objects 2, 4, 6, 8, 
.... marking them 1,2,8,4, . . . . We must then invent 
marks for the objects previously marked 1,3,5,7 The 



NUMERATION AND NOTATION. 19 

marks invented are shown in Figure 1, above the objects, the 
old marks being below the objects. 

041f2 "|3}4J5 

xxxxxxxxxxxx.... 

0123456789 

Fig. i. 

From this it is clear that instead of re-labeling the alternate 

objects in a row of objects, 0, 1, 2, 3, 4, 5, 6, 7, we 

can interpolate alternate objects in the row and then mark them 
J, f , f , and so on. 

In the same way we can interpolate two objects between 

every consecutive two of the given row 0, 1, 2, 3, 4, 5, 

marking the new objects in order J, ; f, |; -J , -f ; and so on. 

4 f 1 | | 2 | | 3 

Thus, XXXXXXXXXX 

0123 

Fig. 2. 

-t n o 

In this way we account for the symbols > ... where 

p P . P 

p is any positive integer. These we call positive fractional 
numbers. 

By interpolating single objects in the row 0, \, 1, f , 2, f , . . . 
we have the same sequence of objects as if we interpolate ob- 
jects by threes in the row 

0, 1, 2, 3, 4, 5, 6, ...... 

and the objects are therefore marked 

0, i, i, i, 1, 

OJi|lffJ2 

Thus, XXXXXXXXXX....^.. 

\ 1 I 2 

Fig. 3. 

From this we see that \ and f are marks for the same object. 
Also | and f . Hence, }=4 and |=f . 

A row marked 0, \, f , J, 4, f , 1, . . . . is to be understood as 
arising from the interpolation of objects by fives; that is, by 
introducing the objects , f , f, f, f ,...., or , J, \, f , f, . . . . 

As f comes before f , we say f<f , or J<i. 

We may interpolate as many objects as we please in the nat- 
ural row, and, by the principle of the least common denomina- 
tor, we can interpolate so as to explain any assigned positive 
fractional marks, /i,/ 2 '/3 Also, given any positive rational 
mark, r, other than zero, we can interpolate rational marks be- 



20 FINKEL'S SOLUTION BOOK. 

tween and r. When no object can be made to fall between an 
assigned object and 0, that assigned object must be itself. 

In the same way we may treat the negative numbers. 

We can think of an infinity of objects as being interpolated 
in the natural row, so that each shall bear a distinct rational 
number and so that we can say which of any two objects comes 
first. It is to be noticed that as we approach any of the natural 
objects there is no last fractional mark; that is, whatever object 
we take there are always others between it and the natural object. 

Thus, if an infinitude of objects be interpolated in the nat- 
ural row, 0, 1, 2, 3,'. . . . 

* i HI 

XXXX . . to in finityXXX . . to infinityXXXXX . . to infinityXX . . to infinityX 
Fig. 4. 1 

then it is clear that whatever object we take there is an infinity 
of objects between it and the natural object, thus rendering it 
evident that there are no last fractional marks in this case. 

12. Irrational Numbers. In considering square num- 
bers from the ordinal point of view, we re-label our natural row 
as in Fig. 5. 

1 i f 2 3 

xxxxxxxxxxxxxxxx 

1 i 2. } 3 4 56789 

Fig. 5. 

where the old names are below and the new above. We have 
now to consider how to bring the omitted objects into the scheme 
of ordinal numbers. Bvery object whose new name is fractional 
had a fractional name, so that the object whose old name was 2 
cannot now have a rational name. We give it a name which we 
call irrational. _We call it the positive or chief square root of 2 
and mark it V '2 or 2*. As an ordinal number it is perfectly 
satisfactory, for we know where it comes, whether left or right 
of any proposed rational number, by means of the old marking. 
Hence, it separates all the rational numbers into two classes, 
viz., those on its right and those on its left. A rational number 
separates all other rational numbers into two classes; we put it 
into one of the classes and say it closes that class. 

Take, for example, f . Now, there is no last fractional mark 
as we approach f from the left or from the right. Hence, with- 
out % neither the class to the left of f nor the class to the right 
of % is closed. With f , either class is closed. 

Any process which serves to separate rational numbers into 
two classes, those on the left and those on the right, such that 
the left-hand class is not closed on the right and the right-hand 
class not closed on the left, leads to the introduction of a new 
object named by an irrational number. 






NUMERATION AND NOTATION. 21 

For example, V 2 separates all rational numbers into two 
classes, viz., those on the left of it and those on the right. Now 
if we take any rational object however near to the V 2 as we 
please ^we can always interpolate new rational objects between it 
and }/~2. Thus, it is clear that the class on the left of i/~2~is 
not closed at the right nor the class on the right closed on the left. 

Two rational or irrational numbers, for simplicity take them 
both irrational and equal to s and /, are equal if the rational 
objects to the left of 5- are the same as the rational objects to the 
left of /, and the rational objects to the right of s are the same 
as those to the right of s r . Thus, 4^ and 2^ effect the same sep- 
aration of the rational numbers. Hence, 4^=2 M . 

An equivalent condition for the equality of ^ and / is that 
every rational number to the left of s shall be to the left of /, and 
every rational number to the left of s' shall be to the left of s. 

Between two unequal irrational objects, s and /, there must 
lie rational objects; for, since s and s' are not equal, there must be 
a rational number which is before one and not before the other. 

It is very important to notice that we have now a closed 
number-system. When we seek to separate the irrational objects 
as lying left or right of an object, either the object is rational or 
if not it separates rational objects and is irrational; in any case 
it must have for its mark a rational or irrational number, and 
there is no loop-hole left for the introduction of new real num- 
bers which separate existing numbers. This is often briefly 
expressed by saying that the whole system of positive and neg- 
ative integral, fractional, and irrational numbers is continuous, 
or is a continuum*. 

In the way indicated above, the number-concept of Arithme- 
tic is put on a basis consistent with Geometry. If we select any 
point on a straight line and call it the zero-point, and also a fixed 
length, measured on this line, be chosen as the unit of length, 
any real number, a, can be represented by a point on this line at 
a distance from the zero-point equal to a units of length. Con- 
versely, each point on the line is at a distance from the origin 
equal to a units of length, when a is a real number. That is, 
there is a one to one correspondence between the points of line 
and the numbers of the real number-system. For every point 
of the line, there corresponds a number of the real number-sys- 
tem and for every number of the real number-system there cor- 
responds a point of the line. 

EXAMPLES. 

1 . Write three hundred seventy quadrillion, one hundred one 
thousand one hundred thirty-four trillion, seven hundred eighty- 

*See Harkness and Morley's Introduction to the Theory of Functions, Chapter I., 
from which this has been adapted. 



22 FINKEL S SOLUTION BOOK 

nine thousand six hundred thirty-two billion, two hundred ninety- 
eight thousand seven hundred sixty-five million, four hundred 
thirty-seven thousand one hundred fifty-six. 

2. Read by the English method, 78943278102345789328903- 
24678. 

3. Write three thousand one hundred forty -one quintillion, 
five hundred ninety-two billion six hundred fifty-three million 
five hundred eighty- nine thousand seven hundred ninety- three 
quadrillion, two hundred thirty -eight billion four hundred sixty- 
two million six hundred forty-three thousand three hundred 
eighty-three trillion, two hundred seventy-nine billion five hun- 
dred two million, eight hundred eighty*-four thousand one hundred 
ninety-seven. 

4. Read 141421356237309504880168872420969807856971437- 
89132. 

5. Is a billion, a million million? Explain. 

6. Write 19 billion billion billion. 

7. Write 19 trillion billion million million. 

8. Write 19 hundred 56 thousand. 

9. Write 457 thousand 341 million. 

10. Write 19 trillion trillion billion billion million million. 



CHAPTEK III. 

ADDITION. 

1. Addition, is the process of uniting two or more numbers 
of the same kind into one sum or amount. 

2. Add the following, beginning at the right, and prove the 
result by casting out the 9's : 

7845 excess of 9=6") 

6780 " " 9=3 

8768 " " 9=2 >Excess of 9's=8. 

5343 " " 9=6 

3987 " " 9=OJ 

32723 excess of 9=8 

Explanation. Ad'ding the digits in the first number, we 
have 24. Dividing by 9, we have 6 for a remainder, which is the 
excess of the 9's. Treating the remaining numbers in the same 
manner, we obtain the excesses 3, 2, 6, 0. Adding the excesses 
and taking the excess of their sum, we have 8 ; this being equal 
to the excess of the sum the work is correct. 



3. 



SUBTRACTION. 

Add the following, beginning at the left : 

8456 
9799 
4363 
5809 
5432 



23 




33859 

From this operation, we see that it is more convenient to be- 
gin at the right 

Remark. We can not add 8 apples and 5 peaches because 
we can not express the result in either denomination. Only 
numbers of the same name can be added. 

EXAMPLES. 

1. Add the numbers comprised between 20980189 and 
20980197. 

2. 6095054 + 900703+90300420+9890655+37699+29753 = 
what? 

3. Add the following, beginning at the left: 97674; 347- 
893; 789356; 98935679; 123456789. 

4. Add all the prime numbers between 1 and 107 inclusive. 

5. Add 31989, 63060, 132991, 1280340, 987654321, 78903, and 
prove the result by casting out the 9's. 

6. Add the consecutive numbers from 100 to 130. 

7. Add the numbers from 9897 to 9910 inclusive. 

8. Add MDCCCLXXVI, MDCXCVIII, DCCCCXLIX, 
DCCCLXII. 



CHAPTER IV. 

SUBTRACTION. 

1. Subtraction is the process of finding the difference be- 
tween two numbers. 

2. Subtract the following and prove the result by casting 
out the 9's : 

984895 excess of 9's=7 
795943 " " 9's=ly 

188952 " " 9'i 



, ^Excess of 9's=7. 
-6 



24 FINKEL'S SOLUTION BOOK. 

Explanation. Adding the digits in the first number, we 
have 43. Dividing by 9 the remainder is 7, which is the excess 
of the 9's. Treating the subtrahend and remainder in the same 
manner, we have the excesses 1 and 6. But subtraction is the 
opposite of addition and since the minuend is equal to the sum 
of the subtrahend and remainder, the excess of the sum of the 
excesses in the subtrahend and remainder is equal to the excess 
in the minuend. This is the same proof as that required if we 
were to add the subtrahend and remainder. 

3. We begin at the right to subtract, so that if a figure of 
the subtrahend is greater than that corresponding to it in the 
minuend, we can borrow one from the next higher denomination 
and reduce it to the required denomination and then subtract. 

4. Subtract the following and illustrate the process : 

1=9 99999 9+1 1=9 9999 9+1 ) A , , 1=9 9 9+1 ) . , 

90000000 9856342 j- Add - 4326546 \ Add - 

85784895 8978567 3214957 



4215105 877775 11 11589 
EXAMPLES. 

1. From 9347893987 take 8968935789. Prove the result by 
casting out the 9's. 

2. 7847893578 6759984699 what ? 

Which is the nearer number to 920864; 1816090 or 27497? 

4. 345673451 8 + 3 -2 + 34 + 7 + 18567 + 43812 1326 4 
678=what. ? 

5. 5 + 6 + 71213 + 1423 + 786 + 5 + 12 8 what \ 

6. 3+4 (6 + 7) 8 + 27 (1 + 3 2 3) (7 8 + 5)3 + 7=* 
what? 



CHAPTER V. 

MULTIPLICATION. 

1. Multiplication is the process of taking one number as 
many times as there are units in another; or it is a short method 
of addition when the numbers to be added are equal. 

2. Multiply the following and prove the result by casting 
out the 9's : 

7855 excess of 9's=7 
435 " " 9's=3 

39275 21 excess of 9's=3. 

23565 
31420 

3416925=excess of 9*8=3. 



MULTIPLICATION. 25 

Explanation. Adding the digits in the multiplicand and 
dividing the sum by 9, the remainder is 7 which is the excess of 
the 9's. Adding the digits in the multiplier and dividing the sum 
by 9, we have the remainder 3 which is the excess of the 9's. 
Now, since multiplication is a short method of addition when 
the numbers to be added are equal, we multiply the excess in the 
multiplicand by the excess in the multiplier and find the excess, 
and this being equal to the excess in the product, the work is 
correct. 

3. Multiply the following, beginning at the left : 

75645 
765 

1st.. 



2d 
3d 




57868425 

3. From this operation, we see that it is more convenient to 
begin at the right to multiply. 

5. In multiplication, the multiplicand may be abstract, or 
concrete; but the multiplier is always abstract. 

6. The sign of multiplication is X > and is read, multiplied by, 
or times. When this sign is placed between two numbers it de- 
notes that one is to be multiplied by the other. In this case, it 
has not been established which shall be the multiplicand and 
which the multiplier. Thus 8x5=40, either may be considered 
the multiplicand and the other the multiplier. If 8 is the mul- 
tiplicand, we say, 8 multiplied by 5 equals 40, but if 5 is the 
multiplicand we say, 8 times 5 equals 40. 

EXAMPLES. 

1. 562402 X345728=what? 

2. 1 mile = 63360 inches; how many inches from the earth 
to the moon the distance being 239000 miles? 

3. Multiply 789627 by 834, beginning at the left to multiply. 

4. 1 acre = 43560 sq. in.; how many square inches in a field 
containing 427 acres? 



26 FINKEL'S SOLUTION BOOK. 

5. Multiply 6934789643 by 34789. Prove the result by cast- 
ing out the 9's. 

6. 2778588 X 34678=what ? 

7. 2X3X4 3x7+3 2X2+4+8X2+4 3 X5+27=what? 

8. 5X7+6X7+8X7 4X6+6 X 6+7 X6=what? 

9. 356789 X4876=what? 

10. 395076 X 576426=what ? 

11. 7733447 X998800=what? 

12. 5654321X999880 what? 



CHAPTER VI. 

DIVISION. 

1. Division, is the process of finding how many times one 
number is contained in another; or, it is a short method of sub- 
traction when the numbers to be subtracted are equal. 

2. Divide the following and prove the result by casting out 
the 9's: 

67)5484888(81864 
536 

Dividend 

124 5484888 excess of 9's=0. 

67 

Quotient 

578 81864 excess of 9's=01 Excess of 9's 

536 U n this product 

Divisor i ,-. 

428 67 excess of 9's=4j equals 0. 

402 



268 
268 

Explanation. Adding the digits in the dividend and di- 
viding the sum by 9, we have the remainder 0, which is the ex- 
cess of the 9's. Adding the digits in the quotient and dividing 
the sum by 9, we have the remainder 0, which is the excess of 
the 9's in the quotient. Adding the digits in the divisor and 
dividing the sum by 9, we have the remainder 4, which is the 
excess of the 9's in the divisor. Since division is the reverse of 
multiplication, the quotient corresponding to the multiplicand, 
the divisor to the multiplier, and the dividend to the product, we 
multiply the excess in the quotient by the excess in the divisor. 
The excess of this product is 0. This excess being equal to the 
excess of the 9's in the dividend, the work is correct. 



DIVISION. 27 

If there be a remainder after dividing, find its excess and add 
it to the excess of the product of the excesses of the quotient and 
divisor. Take the excess of the sum and if it is equal to the ex- 
cess of the dividend the work is correct. 

3. The sign of division is -J-, and is read divided by. 

4. When the divisor and dividend are of the same denomina- 
tion the quotient is abstract ; but when of different denomina- 
tions, the divisor is abstract and the quotient is the same as the 
dividend. Thus, 24 ct. -Met. = 6, and 24ct.-^4 = 6 ct. 

Remark. We begin at the left to divide, that after finding 
how many times the divisor is contained in the fewest left-hand 
figures of the dividend, if there be a remainder we can reduce it 
to the next lower denomination and find how many times the 
divisor is contained in it, and so on. 

Note. The proof by casting out the 9's will not rectify 
errors caused by inserting or omitting a 9 or a 0, or by interchang- 
ing digits. 

EXAMPLES. 

1. 4326422-f-961=what? Prove the result by casting out 
the 9's. 

2. 245379633477-r-1263=what? Prove the result by casting 
out the 9's. 

3. What number multiplied -by 109 with 98 added to the 
product, will give 106700? 

4. The product of two numbers is 212492745 ; one of the 
numbers is 1035; what is the other number? 

5. 27-f-9 X 3-7-9 1+3-^3x9 8-:-4+5X 23 X2-T-2---3-- 
(3X4-J-6+5 2)+81-r-27x3-f-9Xl8-6= what? [Hint. Per- 
form the operations indicated by the multiplication and division 
signs in the exact order of their occurrence.] 

6. (64-4-32X96-7-12 75+3) Xj[(27-:-3)-f-9 1+2] + 
93H-13X7 45 } X9+45-T-9+3 l=what?. 

7. 2x2-r-2-4-2-7-2x2X2-r-2-r-2-r-2=what? Ans. . 

8. 3-r-3-f-3x3x3-7-3-r-OX4x4x5x5=what? Ans. oo. 

9. 2x2x2-f-2x2-7-2-r-2X2X2XOX2X2=what? Ans. 0. 



28 



FINKEL'S SOLUTION BOOK. 



CHAPTER VII. 

COMPOUND NUMBERS. 

1. A Compound Number is a number which expresses 
several different units of the same kind of quantity. 

2. A Denominate Number is a concrete number in 
which the unit is a measure; as, 5 feet^ 7 pints. 

3. The Terms of a compound number are the numbers of 
its different units. Thus, in 4 bu. 3 pk. 7 qt. 1 pt, the terms 
are 4 bu. and 3 pk. and 7 qt. and 1 pt. 

4. Reduction of Compound Numbers is the process 
of changing a compound number from one denomination to an- 
other. There are Two Cases, Reduction Descending" and Re- 
duction Ascending. 

5. Reduction Descending is the process of reducing a 
number from a higher to a lower denomination. 

6. Reduction -Ascending is the process of reducing a 
number from a lower to a higher denomination. 

Ex. Reduce 2 E. Fr. 1 E. En. 2 E. Fl. 3 yd. 2 na. to nails. 




E. Fr. E. En. E. Sc. E. Fl. yd. qr. na. 
21 232 

65 34 



12qr. 5qr. 



6qr. 12 qr. 
6" 
5" 



35 qr. 
4 

140 na. 

2" 

I42na. 



TIME MEASURE. 29 



TIME MEASURE. 

I. Time is a measured portion of duration. 

2. The measures of time are fixed by the rotation of the earth 
on its axis and its revolution around the sun. 

3. A Day is the time of one rotation of the earth on its 
axis. 

4. A. Year is the time of one revolution of the earth 
around the sun. 

TABLE. 

60 seconds (sec.) make 1 minute (min.) 

60 minutes " 1 hour (hr.) 

24 hours " 1 day (da.) 

7 days " 1 week (wk.) 

4 weeks " 1 lunar month (mo.) 

13 lunar months, 1 da. 6 hr. " 1 year (yr.) 

12 calender months ' 1 year. 

365 days " 1 common year. 

365 da'. 5 hr. 48 min. 46.05 sec. " 1 solar year. 

365 da. 6 hr. 9 min. 9 sec. " 1 sidereal year. 

365 da. 6 hr. 13 min. 45.6 sec. " 1 Anomalistic year. 

366 days " 1 leap year, or bissextile year. 
354 days , " 1 lunar year. 

19 years " 1 Metonic cycle. 

28 years " 1 solar cycle. 

15 years " 1 Cycle of Indiction. 

532 years 1 Dionysian Period. 

5. The unit of time is the day. 

6. The Sidereal Day is the exact time of one rotation 
of the earth on its axis. It equals 23 hr. 56 min. 4.09 sec. 

7. The Solar Day is the time between two successive 
appearances of the sun on a given meridian. 

8. The Astronomical Day is the solar day, begin- 
ning and ending at noon. 

9. The Civil Day 9 or Mean Solar Day 9 is the average 
of all the solar days of the year. It equals 24 hr. 3 min. 
56.556 sec. 

1O. The Solar Year, or Tropical Year 9 is the time 
between two successive passages of the sun through the vernal 
equinox. 

II. The Sidereal Year is the time of a complete revolu- 
tion of the earth about the sun, measured by a fixed star. 

13. The Anomalistic Year is the time of two suc- 
cessive passages of the earth through its perihelion. 

13. A Lunar Year is 12 lunar months and consists of 
354 day. 



30 FINKEL'S SOLUTION BOOK. 

14. A Metonic Cycle is a period of 19 solar years, after 
which the new moons again happen on the same days of the year. 

15. A Solar Cycle is a period of 28 solar years, after 
which the first day of the year is restored to the same day 
of the week. To find the year of the cycle, we have the fol- 
lowing rule: 

Add nine to the date, divide the sum by twenty -eight; the 
quotient is the number of cycles, and the remainder is the year of 
the cycle. Should there be no remainder the proposed year is the 
twenty-eighth, or last of the cycle. The formula for the above 

rule is <! \ in which x denotes the date, and r the re- 

mainder which arises by dividing x-\- by 28, is the number 
required. 

Thus, for 1892, we have (1 892+9 )-^28=67ff ' 1892 is the 
25th year of the 68 cycle. 

16. The Lunar Cycle is a period of 19 years, after which 
the new moons are restored to the same day of the civil month. 

The new moon will fall on the same days in any two years 
which occupy the same place in the cycle; hence, a table of the 
moon's phases for 19 years will serve for any year whatever 
when we know its number in the cycle. This number is called 
the Golden Number. 

To find the Golden Number : Add 1 to the date, divide the 
sum by 19; the quotient is the number of the cycle elapsed and the 
remainder is the Golden Number. 

r*+ii 

The formula for the same is <j \ in which r is the re- 

mainder after dividing the date-}-l by 19. It is the Golden 
Number. 

17. A Dionysian or Paschal Period is a period of 
532 year, after which the new moons again occur on the same 
day of the month and the same day of the week. It is obtained 
by multiplying a Lunar Cycle by a Solar Cycle. 

18. A Cycle Of Induction is a period of 15 years, at 
the end of which certain judicial acts took place under the Greek 
emperors. 

19. JEpact is a word employed in the calender to signify 
the moon's age at the beginning of the year. 

The common solar year, containing 365 days, and the lunar 
year only 354, the difference is 11 days; whence, if a new 'moon 
fall on the first of January in any year, the moon will be 11 days 
old on the first day of the following year, and 22 days on the 
first of the third year. The epact of these years are, therefore, 
eleven and twenty-two respectively. Another addition of eleven 



LONGITUDE AND TIME, 



31 



lays would give thirty-three for the epact of the fourth year; but 
in consequence of the insertion of the intercalary month in each 
third year of the lunar cycle, this epact is reduced to three. In 
like manner the epacts of all the following years of the cycle are 
obtained by successively adding eleven to the epact of the former 
year, and rejecting thirty as often as the sum exceeds that 
number. 

LONGITUDE AND TIME. 

In the diagram, the curve ACBD represents the path of the 
earth in its journey around the sun. This curve is called an 
ellipse. The ellipse may be drawn by taking any two points 
.Fand F' and fastening in them the extremities of a thread whose 




length is greater than the distance F' F. Place the point of a 
pencil against the thread and slide it so as to keep the thread 
constantly stretched ; the point of the pencil in its motion will 
describe the ellipse. 

The points F and F' are called the/0, the plural of focus. 

The sun occupies one of these foci. The plane of the earth's 
orbit, or path, is called the ecliptic. When the earth is at A it is 
nearer the sun than when it is at B. When the earth is nearest 
the sun it is said to be in perihelion (Gr. Kepi=peri, near, and 
fytos=halios, sun); when farthest from the sun, it is said to be 
in aphelion (Gr. dn6=apo, from, and r\kw~-=halios, sun). The 
points A and B, in the diagram, represent the perihelion and 
aphelion distances, respectively. The earth is nearest the sun 
about the first of January and farthest from the sun about the 
first of July. It takes the earth 365 da. 6 hr. 13 min. 45.6 sec. to 
travel from A, west around through C, B, and D back to A. 



32 FINKEL'S SOLUTION BOOK. 

This period of time is called the anomalistic year. The west 
point as here spoken of, may be thought of thus : Suppose you 
were located at some point on the surface of the sun in a posi- 
tion enabling you to see the North Star. Then if you should 
face that star you would be facing north, your right hand would 
be to the east, and your left hand to the west, and south to 
your back. 

While the earth makes one revolution around the sun, it 
rotates 366 times on one of its diameters. The diameter upon 
which it rotates is called its axis. The axis of the earth is in- 
clined from a perpendicular to the plane of the earth's orbit at 
an angle of 28J. If the axis of the earth were extended indefi- 
nitely, it would pass very near, 1J, from the North Star. 

The earth's axis and the sun determine a plane, and this 
plane is of great importance in gaining a thorough understand- 
ing of Longitude and Time. Suppose you are on the earth's 
surface, facing the sun and in this plane. Then you will have 
noon, while just to the east of you it will be after noon and just 
to the west it will be before noon. The intersection of this plane 
with the earth's surface is called the trace of the plane on the 
earth's surface. This trace is called a meridian. 

If you could travel in such a way as to remain in this plane 
for a whole day, that is, 24 hours, you would have noon during 
the whole time. But if you remain stationary on the earth's 
surface, you will be carried out of this plane eastward by the 
earth's rotation. You may conceive that you are at Greenwich, 
formerly a small suburban town of London, Eng., but now in- 
corporated in that city, with the sun visible and having such a 
position that you are in the plane formed by the earth's axis and 
the sun. It will then be noon to you at that place. 

Suppose we take the trace of the plane, in this position, on 
the earth's surface as our standard meridian. Then all places 
east of this line will have had noon and all places west of it 
are yet to have noon. As the earth continues to rotate, rotating 
as it does from west to east, it will bring points west of the 
plane into coincidence with the plane and thus these points will 
have noon successively as they come into the plane. Suppose 
we start when Greenwich is in this plane, and mark the trace of 
the plane on the earth's surface and every four minutes we mark 
the trace of the plane; in this way, in a complete rotation of the 
earth, we will have drawn 360 of these traces, which we have 
agreed to call meridians. 

The distance of these lines apart, measured on the equator, 
is called a degree of longitude, better an arc-degree of longitude. 
Instead of measuring longitude from Greenwich entirely around 
the earth through the west, we generally measure it east and 
west to 180. 



LONGITUDE AND TIME. 33 

Thus, a place located on the 70th meridian, west, is said to 
be 70 west longitude, and a place situated on the 195th merid- 
ian, counting from Greenwich around through the west, is said 
to be 85 east longitude. 

From the above discussion, we see that, since the earth turns 
on its axis once in 24 hours, 

24 hrs. =360 of long., or 360 of long.=24 hrs. 
1 hr. =^3- of 360=! 5 of long., or 15 of long.=l hr. 
1 min.= g J - of 15=15' of long., or 15' of long.=l min. 
1 sec. =fa of 15' =15" of long., or 15" of long.=l sec. 

Hence, if we have the difference of longitude of two places, we 
can readily find the difference of time between these two places. 

For example, the longitude of St. Petersburg is 30 16' E., 
and the longitude of Washington is 77 0' 36" W. Now the dif- 
ference of longitude between these two places is 77 0' 36" + 
30 l&=m 16' 36". Hence, since 15=1 hour, 107 16' 36"= 
(107 16' 36")-KL5 f or 7 hrs. 9 min. 6.4 sec., which is the differ- 
ence of time between Washington and St. Petersburg. 

Conversely, if we know the difference of time between two 
places, we can easily find the difference of longitude. 

For example, the difference of time between New York City 
and St. Louis is 1 hr. 4 min. 47J sec. Find the difference of 
longitude. 

1 hr. =15. 
1 min. =15'. 
II. 1 3. 4 min. =60', or 1. 

Isec. =15". 
5. 47J- sec. =47JX15"=710"=11' 50". 

III. Hence, the difference of longitude is 16 11' 50". 

In some cases, problems are so proposed that we are to find 
the longitude or time of one place, having given the longitude 
or time of another place and the difference of time or difference 
of longitude of the two places. Such problems require no prin- 
ciples beyond those already established. 

For example, the difference of time between two places is 
2 hr. 30 min. The longitude of the eastern place is 56 W. 
Find the longitude of the western place. 

1. Ihr. =15. 

2. 2hr. =30. 

3. 1 min.=15'. 

II. <! 4. 30 min.=30X15'=450=7 30'. 

5. .'. 37 30'=difference of longitude. 

6. .'. 56+37 30'=93 30', the longitude of the west- 

ern place. 



34 FINKEL'S SOLUTION BOOK. 

Had the place whose longitude is given been in east longi- 
tude, we would have subtracted the difference of longitude to 
find the longitude of the western place. 

The following suggestions may prove helpful in the solution 
of problems in Longitude and Time : 

1. When the longitude of a place is required, having given 
the longitude of some other place and the difference of longitude 
between the two places. 

Conceive yourself located at the place whose longitude is 
given. Then ask yourself this question, Is the place whose 
longitude is required, east or west? If west, add the difference 
of longitude when the given longitude is west, and subtract if 
the given longitude is east. If the answer to your question is 
east, subtract the difference of longitude when the given longi- 
tude is west and add the difference of longitude when the given 
longitude is east. 

If the places are on opposite sides of the standard meridian, 
subtract the given longitude from the difference of longitude and 
the difference will be the longitude required, and will be oppo- 
site in name from the given longitude. That is to say, if the 
given longitude is east, the required will be west, and vice versa. 

2. When the time of place is required, having given the time at 
some other place and the difference of time between the two places. 

Conceive yourself located at the place whose time is given. 
Then ask yourself this question, Is the place whose time is 
required, east or west of me? If the answer to your question 
is west, subtract the difference of time for the required time. 
If the answer to your question is east, add the difference of 
time for the required time. 

STANDARD TIME. 

In 1883, the railroad officials of the United States and Canada 
adopted what is called standard time. These officials agreed to 
adopt the solar time of some standard meridian as the local time 
of an extended area. The standard meridians thus adopted are 
75th, 90th, 105th, and 120th. All stations in the belt of country 
7^- wide on either side of these standard meridians have as local 
time the solar time of the respective meridian. For example, 
all points or stations in the belt of country 7J wide on either side 
of 90th meridian, i. e., the belt of country lying between 82^ 
and 97-| west longitude have as local time the solar time, or sun 
time, of the 90th meridian. In other words, all time-pieces of 
the various stations in this belt indicate the same time of day as 
clocks in the depots situated on the 90th meridian. For exam- 
ple, when the clock m the Union Depot at St. Louis indicates 
noon, 12 o'clock M., the clocks in the Union Depots at Indian- 



STANDARD TIME. 35 

apolis and Kansas City also indicate noon, though at Indianap- 
olis it is a little more than 16 min. past noon and at Kansas City 
it is a little more than 17 min. till noon, sun time. That is, the 
standard time and local time at Indianapolis differ by a little 
more than 16 min., standard time being about 16 min. slower 
than local time, and at Kansas City standard time and local time 
differ by about 17 miu., standard time being about 17 min. faster 
than local time. 

If one were to set his watch with the railroad clock in the 
depot at Columbus, Ohio, then take the train for Springfield, 
Mo., on arriving at Springfield, Mo., one would find that his 
watch agrees with the clock in the Frisco depot. This is be- 
cause Columbus, Ohio, and Springfield, Mo., are located in the 
belt of country having central time, i. e., having the sun time of 
the 90th meridian. 

How about the local time of these two places? The local 
time at Columbus, O., is about 28 min. faster than standard time. 
The sun comes to the meridian of Columbus before it comes to 
the 90th meridian. When the sun comes to the meridian at 
Columbus it is noon, local time, but it will not be noon, standard 
time, until the sun comes to the 90th meridian, which will be 
about 28 min. later. Hence, local time at Columbus, O. , is about 
28 min. faster than standard time. A passenger going from 
Columbus, Ohio, to Springfield, Mo., and carrying standard time 
of Columbus, would have standard time at St. L,ouis, standard 
time and local time at St. L<ouis being very nearly the same. On 
arriving at Springfield, Mo., his watch would still indicate stand- 
ard time and would agree with the regulator in the depot at 
Springfield, but his time would be about 8 min. faster than the 
local time at Springfield. 

The time in the belt of country between 67J- west longitude 
and 82^ west longitude is called Eastern Time ; between 82-^- 
W. and 97i W., Central Time; between 97^- W. and 112 W., 
Mountain Time; and between 11 2^ W. and 127J W., Pacific 
Time. We might call the time in the belt between 7^ E. and 
7 W,. Greenwich Time; between 7J W. and 22| W., East 
Atlantic Time; between 22^ W. and 37^ W., Central Atlantic 
Time; between 37J W. and 52^ W., West Atlantic Time; and 
between 52^ W. and 67J C W., ^Colonial Time. 

By some appropriate system of nomenclature, the naming of 
the time in the belt beginning with 127^ W. longitude might 
be extended. However, these names would have a very limited 
use and are therefore not worth coining. 



36 FINKEIv'S SOLUTION BOOK. 



THE INTERNATIONAL DATE LINE. 

The International Date Line is an irregular line pass- 
ing through Bering Strait, along the coast of Asia to near 
Borneo and Philippine Islands, and thence along the northern 
limits of the East Indian Islands, New Zealand, and New Guinea. 
It is the line from which every date on the earth is reckoned. 
At present, however, the 180th meridian is very generally used 
in its stead. 

Suppose one were standing on the 180th meridian at the time it is 
noon, Wednesday (say), at Greenwich, and facing north. Then, just to 
right, or east of this line, Wednesday is beginning, i. e., Wednesday 12 
o'clock, A. M M while just to left, or west of the line, Wednesday is ending, 
i. e., Wednesday 12 o'clock, P. M. The difference of time between places 
immediately east and immediately west of the line is therefore 24 hours, 
and just west of the line it is one day later than just east of it. This is 
made still clearer by considering what takes place as the earth rotates on 
its axis; the places just west of the line will be carried eastward, and since 
these places had Wednesday ending, they must now have Thursday begin- 
ning. But these places are west of the line, the places east of the line 
still having Wednesday. Hence, it is clear that it is one day later just 
west of the line than just east of the line. In crossing this line, there- 
fore, from the east one day must be added, while in crossing it from the 
west one day must be subtracted. 

Professor C. A. Young, in his General Astronomy, in answering the 
question, "Where does the day begin?" says, "If we imagine a traveler 
starting from Greenwich on Monday noon and traveling westward as swiftly 
as the earth turns to the east under his feet, he would, of course, keep the 
sun exactly on the meridian all day long and have continual noon. But 
what noon ? It was Monday when he started, and when he gets back to 
Ivondon, twenty-four hours later, it is Tuesday noon there, and there has 
been no intervening sunset. When does Monday noon become Tuesday 
noon? The convention is that the change of date occurs at the 180tb 
meridian from Greenwich. A ship crossing this line from the east skips 
one day in so doing. If it is Monday forenoon when the ship reaches the 
line, it becomes Tuesday forenoon the moment it passes it, the intervening 
twenty-four hours being dropped from the reckoning on the log-book. 
Vice versa, when a vessel crosses the line from the western side, it counts 
the same day twice, passing from Tuesday forenoon back to Monday, and 
having to do its Tuesday over again." 

This line is now little used by sailors, the 180th meridian 
having taken its place. 

The consideration of this line in the solution of problems in 
longitude and time should add no serious difficulty. Solve the 
problem completely, leaving out of account the date line. Then, 
if the time of the given place is west of the line, while the place 
whose time is required is east, we simply subtract a day, and if 
the conditions are reversed, we add a day. 

I. When it is five minutes after four o'clock on Sunday 
morning at Honolulu, what is the hour and day of the week at 
Sydney, Australia? (Ray's Higher Arithmetic, p. iji,prob. 7.) 



THE INTERNATIONAL DATE LINE. 



37 



II. < 



1. 



x HONOLULU 



157 52 W.= 
longitude of 
Honolulu. 



OREEWU:M^*-T:- 
^3^ 



151 11 E.= 
longitude of 
Sydney. 

309 3'=differ- 
ence of long- 
itude meas- 
ured from 
Honolulu 
through 
Greenwich 
to Sidney. 

360 309 3' = 50 57'= difference of longitude 
measured directly on the equator from the merid- 
ian through Honolulu to the meridian through 
Sydney. 

15=1 hr. 

1=-- hr.=4 min. 



5. 
6. 

7. 50 57"=50fJr=50i =50^X4 min.=3 hr. 23 min. 

48 sec., difference of time. 

8. 4 hr. 5 min., Sunday 3 hr. 23 min. 48 sec. =41 min. 

12 sec., Sunday. 

9. Regarding the date line, Sunday is changed to Mon- 

day, since Honolulu is east of the line, while Syd- 
ney is west of it. 

EXAMPLES. 

1. When it is 5 o'clock Monday morning at Paris, France, longitude 
2 20' E., what is the hour and day of the week at Honolulu, Hawaiian Is- 
lands, longitude 157 52' W.? 

Ans. 19 min. 12 sec. past 6 o'clock P. M., Sunday. 

2. When it is five minutes after 3 o'clock on Sunday morning at Hon- 
olulu, Hawaiian Islands, longitude 157 52' W., whet is the hour and day of 
the week at Sydney, Australia, longitude 151 II 7 E.? 

Ans. 41 min. and 12 sec. before 12 o'clock P. M., Sunday. 

3. When it is 20 minutes past 12 o'clock on Saturday morning at Chi- 
cago, 111., longitude 87 35', what is the hour and day of the week at Pekin, 
China, longitude 116 26' E.? 

Ans. 56 min. 4 sec. past 1 o'clock P. M., Saturday. 

4. When it is ten minutes until 12 o'clock, Friday, midnight, at Con- 
stantinople, Turkey, longitude 28 59 X E., what is the hour and day of the 
week at Honolulu, Hawaiian [slands, longitude 157 52 X W.? 

Ans. 22 min. 36 sec. past 11 o'clock A. M., Friday. 

5. At what hour must a man start, and how fast must he travel, at the 
equator, so that it would be noon for him for twenty-four hours ? 

Ans. Noon ; 1037.4 statute miles per hr. 



38 



FINKEL'S SOLUTION BOOK. 



6. What is the difference of time between Constantinople, Turkey, 
and Sydney, Australia? Ans. 8 hr. 10 min. 48 sec. 

7. A traveler sets his watch with the time of the sun at New York. 
He then travels from there and on arriving at his destination finds that his 
watch is 1 hr. 20 min. 30 sec. fast. What is the longitude of his destination 
if the longitude of New York is 74 (X 24" W.? Ans. 94 7' 54" W. 

8. When it is 1 o'clock P. M. at Rome, Italy, longitude 12 28' E., what 
is the hour at New York, longitude 74 O x 24" W.? 

Ans. 14 min. G^- sec. past 7 A. M. 

SOLUTIONS. 
Ex. 2. Reduce 2 p. 3 pn. 1 tr. 1 hhd. 1 gal. 1 qt. to pints, 

126 
84 
42 
31* 

63 4 2 
bbl. T. hhd. gal. qt. pt. 

1 11 

63 63 




P- 
2 
126 



pn. 
3 

84 



tr. 
1 
42 



252 gal. 252 gal. 42 gal. 



63 gal. 

42 
252 
252 



610 gal. 
4 



2440 qt. 




4882 pt. 

Ex. 3. I. Reduce 2 bu. 3 pk. 2 qt. 1 pt. to pints. 
Equation Method. 

-1. 1 bu.=4 pk. 
2. 2 bu.=2x4 pk.=8pk. 
8 pk.+3 pk.=ll pk. 

I pk.=8 qt. 

II pk.=llx8qt.=88 qt. 
88 qt.+2 qt.=90 qt. 

90 qt.=90x2 pt.=180 pt. 
180 pt.-f 1 pt.=181 pints. 
.-. 2 bu. 3 pk. 2 qt. 1 pt.=181 pints. 



Solution : II. 



Conclusion : III. 



EXAMPLES. 



Solution: II. 



Conclusion: 



Solution : II. 



Ex. 4. I. Reduce 529 pints to bushels. 
Equation method. 

1. 2 pt.=l qt. 

2. 529 pt.=529-4-2=264 qt.+l pt. 

3. 8 qt.=l pk. 

4. 264 qt.=264-7-8=33 pk. 

5. 4 pk. 1 bu. 

6. 33 pk.=33-f-4=8 bu.+l pk. 
III. .-. 529 pints=8 bu. 1 pk. 1 pt. 

Ex. 5. How many gallons will a tank 4 ft. long, 3 ft 
wide, and 1 ft. 8 in. deep contain? 

'l. 4 ft.=length, 

2. 3 ft.=width, and 

3. 1 ft. 8 in.=l ft.=depth. 

4. 4x3Xlf=20 cubic ft.=contents of tank. 

5. 1 cu. ft.=1728 cu. in. 

6. 20 cu. ft.=20Xl728 cu. in.=34560 cu. in. 

7. 231 cu. in.=l gal. 

8. .;. 34560 cu. in.=34560-f-231=149ff gal. 

Conclusion : III. .-. The tank will contain 149ff gallons. 

(Fish's Comp. Arith., p. 126, prob. 2.) 

EXAMPLES. 

1. How many links in 46 mi. 3 fur. 5 ch. 25 links? 

2. How many acres in afield containing 1377 square chains? 

3. How many cubic inches in 29 cords of wood? 

4. In 1436 nails how many Ell English? 

5. How many miles in 3136320 inches? 

6. In 47 ft). 2 1 3 3 1 3 19 gr. how many grains ? 

7. Change 16 Ib. 3 oz. 1 gr., Troy weight to Avoirdupois 
weight. 

8. An apothecary bought by Avoirdupois weight, 2 ft). 8 oz. of 
quinine at $2.40 per ounce, which he retailed at 20 ct. a scruple. 
What was his gain on the whole? 

9. How many seconds in a Dionysian Period? 

10. How many seconds in the month of February, 1892. 

11. How many seconds in the circumference of a wagon 
wheel? 

12. How long would it take a body to move from the earth to 
the moon, moving at the rate of 30 miles per day. 

13. If a man travels 4 miles per hour, how far can he travel in 
2 weeks and 3 days? 



40 FINKEL'S SOLUTION BOOK. 

14. How much may be gained by buying 2 hogsheads of mo- 
lasses, at 40 ct. per gallon, and selling it at 12 cents per quart? 

Ans. $10.08 

15. In 74726807872 seconds, how many solar years? 

Ans. 2368 years. 

16. At $4 per quintal, how many pounds of fish may be 
bought for $50.24? Ans. 1256 pounds. 

17. How many bottles of 3 pints each will it take to fill a 
hogshead? Ans. 168. 

18. What will 73 bushels of meal cost, at 2 cents per quart? 

Ans. $46.72. 

19. How many ounces of gold are equal in weight to 6 ft), of 
lead? Ans. 87oz. 

20. What is the difference between the weight of 42f ft), of 
iron and 42.375 ft), of gold ? Ans. 52545 gr. 

21. How many bushels of corn will a vat hold that holds 
5000 gallons of water. Ans. 537 A bu. 

22. A cellar 40 ft. long, 20 ft. wide and 8 ft. deep is half full 
of water. What will it cost to pump it out, at 6 cents a 
hogshead ? Ans. $22.797+. 

23. If a man buys 10 bu. of chestnuts at $5 a bushel, dry meas- 
ure, and sells the same at 25 cents a quart, liquid measure, how 
much does he gain? Ans. $43.09-f- gain. 

24. How many steps, 2 ft. 8 in. each, will a man take in walk- 
ing a distance of 15 miles? Ans. 29700. 

25. How many hair's width in a 40 ft. pole, if 48 hair's width 
equals 1 line? 

26. How many chests of tea, weighing 24 pounds each, at 43 
cents a pound, can be bought for $1548? Ans. 150 chests. 

27. How long will it take to count 6 million, at the rate of 80 a 
minute, counting 10 hours a day? Ans. 125 days. 

28. How long will it take to count a billion, at the rate of 80 
a minute, counting 12 hours a day? Ans. - - 

29. What will 15 hogsheads of beer cost, at 3 cents a pint. 

Ans. $194.40. 

30. How many shingles will it take to cover the roof of a 
building 60 ft. long and 56 ft. wide, allowing each shingle to be 4 
inches wide and 18 inches long, and to lie ^ to the weather? 

. 20160. 



31. There are 9 oz. of iron in the blood of 1 man. How many 
men would furnish iron enough in their veins to make a plow- 
share weighing 22-J- Ibs. ? Ans. 40- 



GREATEST COMMON DIVISOR. 41 

CHAPTER VIII. 

GREATEST COMMON DIVISOR. 

1. A Divisor of a number is a number that will exactly 
divide it. 

2. A Common Divisor of two or more numbers is a 
number that will exactly divide each of them. 

3. The Greatest Common Divisor 9 or Highest 
Common Factor, of two or more numbers is the greatest 
number that will exactly divide each of them. 

I. Find the G. C. D. of 60, 120, 150, 180. 

II. 60=2X2X3X5. 
2. 120=2X2X2X3X5. 
3. 150=2X3X5X5. 
4. 180=2X2X3X3X5. 
5. G. C. D.=2X3X5=30. 
III. .-. G. C. D.=30. 

Explanation. By inspecting the factors of each number we 
observe that 2 is found in each set of factors; hence, each of the 
numbers can be divided by 2. But only once, since it is found 
only once in the factors of 150. We also observe that 3 will 
divide the numbers only once, since it occurs only once in the 
factors of 60 and 120. Also, 5 will divide them but once, since 
60, 120 and 180 contain it but once. Hence, the numbers, 60, 
120, 150, 180, being, divisible by 2, 3 and 5, are divisible by their 
product, 2X3X5=30. 

I. Find the G. C. D. of 180, 1260, 1980. 

(1. 180=2X2X3X3X5. 
TT J2. 1260=2X2X3X3X5X7. 
' ]3. 1980=2X2X3X3X5X11- 

U. G. C. D.=2X2X3X3X5=180. 
III. .-. G. C. D. of 180, 1260, 1980=180. 

Explanation. 2 being found twice in each number, they 
are each divisible by 2x2 or 4 ; also 3 being found twice in each 
number, they are each divisible by 3x3 or 9. 5 being found in 
each number, they are each divisible by 5. Hence, they are 
divisible by the product of these factors, 2X2X3X3X5=180. 

EXAMPLES. 

1. Find the G. C. D. of 78, 234, and 468. 

2. What is the G. C. D. of 36, 66, 198, 264, 600 and 720? 

3. I have three fields : the first containing 16 acres, the second 
20 acres, and the third 24 acres. What is the largest sized lots 



42 FINKEL'S SOLUTION BOOK. 

containing each an exact number of acres, into which the whole 
can be divided? Ans. 4 A. lots. 

4. A farmer has 12 bu. of oats, 18 bu. of rye, 24 bu. of corn 
and 30 bu. of wheat. What are the largest bins of uniform size, 
and containing an exact number of bushels, into which the whole 
can be put, each kind by itself, and all the bins be full? 

Ans. 6 bu. bins. 

5. A has a four-sided field whose sides are 256, 292, 384, and 
400 feet respectively; what is the length of the rails used to fence 
it, if they are all of equal length and the longest that can be 
used? Ans. 4 ft. 

6. In a triangular field whose sides are 288, 450, and 390 feet 
respectively, how many rails will it require to fence it, if the 
fence is 5 rails high, and what must be the length of the rails if 
they lap over one foot? Ans. Length of rail, 7 ft. No. 940. 



CHAPTER IX. 

LEAST COMMON MULTIPLE. 

1. A. Multiple of a number is a number that will exactly 
contain it ; thus, 24 is a multiple of 6. 

3. A Common, Multiple of two or more numbers is a 
number that will exactly contain each of them. 

3. The Least Common Multiple of two or more num- 
bers is the least number that will exactly contain each of them. 
I. Find the L. C. M. of 30, 40, 50. 

II. 30=2X3X5. 
2. 402X2X2X5. 
3. 50=2X5X5. 
4. L. C. M.=2X2X2X3X5X5=600. 
III. .-. L. C. M. of 30, 40, 50-=600. 

Explanation. The L. C. M. must contain 2 three times, or 
it would not contain 40; it must contain 5 twice, or it would not 
contain 50; it must contain 3 once, or it would not contain 30. 
Since all the factors of the numbers, 30, 40, 50, are contained in 
th^ L. C. M., it will contain each of them without a remainder. 

I. Find the L. C. M. of 2310, 2,10, 30, 6. 

II. 2310=2X3X5X7X11- 
2. 210=2X3X5X7. 
3. 30=2X3X5. 
4. 6=2X3. 
5. L. C. M.=2X3X5X7X 11=2310. 
III. .-. L. C. M. of 2310, 210, 3(\ 3 



LEAST COMMON MULTIPLE. 43 

Explanation. 2 and 3 must be used, else the L. C. M- 
would not contain 6. 2, 3, and 5 must be used, else the L. C. M- 
would not contain 30. Hence 5 must be taken with the factors 
of 6. In like manner 7 must be taken with the factors already 
taken, else the L. C. M. would not contain 210. The factor 11 
must be taken with those already taken, else the L. C. M. would 
not contain 2310. Hence 2, 3, 5, 7, and 11 are the factors to be 
taken and their product 2310 is the L. C. M. 

I. The product of the L. C. M. of three numbers between 
1 and 100 is 6804 ; and the quotient of the L. C. M. divided by 
the G. C. D. is 84. What are the numbers? 

1. L. C. M.xG. C. D.=6804, and 

L.C.M 

c P TT 
3. .-.'L.'C.'M.XG. C. D.-s-k' C' p=L. C. M. X 

G. C. D -X''=(G- C. D. ) 2 =6804-;-84=81. 



II. 



4. G. C. D. y'si =9, by extracting the square root. 



5. .-. L. C. M.=6804-i-9=756. 

6. 9=3x3. 

7. 756=2x2x3x3x3x7. 

8. 3x3x2x2=36. 

9. 3x3x3x2=54. 
10. 3x3x7 =63. 

III. .-. 36, 54, and 63=the numbers. 

Explanation. Since 9 is the G. C. D., each of the numbers 
contains the factors of 9. Since there are two 2's in the L. C. 
M., one of the numbers must contain these factors. In like man- 
ner one of the numbers must contain three 3's; one of them must 
also contain 7. .' We write two 3's for each of the numbers, two 
2's to any set of these 3's, and 3 and 7 with either of the remain- 
ing sets, observing that the product of the factors in any set does 
not exceed 100. If we omit 2 in step 9, the product of the fac- 
tors is 27. Hence 27, 36, 63 are numbers also satisfying the con- 
ditions of the problem. 

EXAMPLES. 

1. What is the L. C. M. of 13, 14, 28, 39, and 42? 

2. What is the L. C. M. of 6, 8, 10, 18, 20, 36, and 48? 

3. What is the L. C. M. of 18, 24, 36, 126, 20, 48, 96, 720, 
and 84? 

4. What is the smallest sum of money with which I can 
purchase a number of oxen at $50 each, cows at $40 each, or 
horses at $75 each? Ans. $600. 



44 



FINKEL'S SOLUTION BOOK. 



5. Find three numbers whose L. C. M. is 840 and G. C. D. 
42. Ans. 84, 210, and 420. 

6. What three numbers between 30 and 140 having 12 for 
their G. C. D. and 2772 for their L. C. M. ? Ans. 36, 84, and 132. 

7. At noon the second, minute, and hour hands of a clock 
are together; how long after will they be together again for tho 
first time? 

8. J. S. H. has 5 pieces of land; the first containing 3 A. 
2 rd. 1 p.; the second, 5 A. 3 rd. 15 p.; the third 8 A. 29 p. v 
the fourth, 12 A. 3 rd. 17 p.; and the fifth, 15 A. 31 p. Re- 
quired the largest sized house-lots, containing each an exact 
number of square rods, into which the whole may be divided. 

Ans. 1 A. 21 p. 

9. The product of the L. C. M. of three numbers by their 
G. C. D.==864, and the L. C. M. divided by the G. C. D.=24; 
find the numbers. Ans. 12, 18, and 48. 



CHAPTER X. 

FRACTIONS. 
1, A Fraction is a number of the equal parts of a unit. 

1. Simple. 
1. As to Form.<! 2. Complex. 



2. Fraction. < 



'1. 


Common, J 
or 
Vulgar. 


,2. 


As to Value." 


I 

i* 


Proper. 
Improper. 
Mixed. 


2. 


Decimal. J 


'l. 
2. 
3. 


Pure. 
Mixed. I 
Circulating. <^ 




Pure. 
Mixed. 


3. 


Continued 


Fractions. 



3. A Common Fraction, or Vulgar Fraction, is 

one in which the unit is divided into any number of equal parts; 
and is expressed by two numbers, one written above the other, 
with a horizontal line between them. Thus, expresses five- 
sixths. 

4. A Simple Fraction is a fraction having a single 
integral numerator and denominator; as, -f. 

5. A Complex Fraction is a fraction whose numer- 
ator, or denominator, or both, are fractional; as, , ^f, -^. 

of o^ 5 




FRACTIONS. 45 

6. A CoMiptttt'fld Fraction is a fraction of a fraction; 
as, J of f . 

7. A jProper Fraction is a simple fraction whose 
numerator is less than its denominator; as, |. 

8. An Improper Fraction is a simple fraction whose 
numerator is greater than its denominator; as, |. 

9. A Mixed Wumber is a whole number and a frac- 
tion; as, 3|. 

1C. A Decimal Fraction is a fraction whose denomi- 
nator is ten, or some power of ten; as, y 3 ^, T | , T |^. The de- 
nominator of a decimal is usually omitted and the point (.) is 
used to determine the value of the decimal expression. Thus, 



11. A Pure Decimal is one which consists of decimal 
figures only; as, .375. 

13. A Mixed Decimal is one which consists of an 
integer and a decimal; as, 5.25. 

13. A Circulating Decimal, or a Circulate, is a deci- 
mal in which one or more figures are repeated in the same order; 
as, .2121 etc. When a common fraction is in its lowest terms 
and the denominator contains factors other than 2 or powers of 
2, and 5 or powers of 5, the equivalent decimal fraction will be 

7 

circulating. Thus, T^Vir^o^ 5 FT wl ^5 when reduced to a 

L X o X o 

decimal, be circulating because the denominator contains the 
factor 3. 

The repeating figure or set of figures is called a Repetend* 
and is indicated by placing a dot over the first and the last fig- 
ure repeated. 

14. A Pure Circulate is one which contains no figures 
but those which are repeated; as, .273. 



15. A Mixed Circulate is one which contains one or 
more figures before the repeating part; as, .45342. 

16. A Simple Hepetend contains but one figure; as, .3. 

17. A Compound Hepetend contains more than one 
figure; as, 354. 

18. Similar Hepetends are those which begin and end 
at the same decimal places; as, .3467) and .0358- 

19. Dissimilar Repetends are those which begin or 
end at different decimal places ; as, .536. .835, and .3567. 



46 FINKEL'S SOLUTION BOOK. 

2O. A Perfect Hepetend is one which contains as many 
decimal places, less 1, as there are units in the denominator of the 
equivalent common fraction ; thus, 7=.142857- 



21. Conterminous Jiepetends are those which end at 
the same decimal place; as, .4267, -3275, and .0321. 

22. Co-originous Hepetends are those which begin at 
the same decimal place ; as, .378, -5624, and 3-623. 

I. Reduce - to its lowest terms. 



Explanation. Dividing the numerator 9, by 3, without 
changing the denominator, the value of the fraction is dimin- 
ished as many times as there are units in the divisor 3. Dividing 
the denominator 12, by 3, without changing the numerator 9, the 
value of the fraction is increased as many times as there are units 
in the divisor 3. Hence, if we divide both terms by 3, the in- 
crease by dividing the denominator will be equal to the decrease 
by dividing the numerator, and the value of the fraction will 
remain unchanged. 

I. Reduce j- to a higher denomination. 



Explanation. Multiplying the numerator 2, by 4, without 
changing the denominator, the value of the fraction is increased 
as many times as there are units in the multiplier 4. Multiply- 
ing the denominators, by 4, without changing the numerator, the 
value of the fraction is decreased as many times as there are 
units in the multiplier 4. Hence, if we multiply both terms by 
4, the increase by multiplying the numerator is equal to the de- 
crease by multiplying the denominator, and the value of the 
fraction remains unchanged. 

I. Reduce 9 to an improper fraction. 

i. 9 



oi* TT J 2 - l=i=8-eighths. 

1 3. 9=9Xf =-V-=9X8-eighths=r72-eighths. 

Conclusion- III. .'. 9%=^j>-=79-eighths. 

I. Reduce f to 24ths. 

' 1. f =f f , or 8-eighths=24-twenty-fourths. 

2. 4=4 of }J=A. or l-eighth=4 of 24- 
twenty-fourths=3-twenty-fourths. 

f :=5 times /j:^^! , or 5-eighths=5 times 
3-twenty-fourths=15-twenty-fourths. 
|=:||=r!5-twenty-fourths. 



II. } 
3. 



FRACTIONS. 



47 



I. Reduce f to 8ths. 

1. f=|, or 6-sixths=:8-eighths. 

9 i i ~f 8 f .a li, or 

* t 01 f o fi- 



ll. 



III. 
I. 



3. 4=5 times ?=- 



=J- of 8- 

eighths. 

or 5-sixths=5 times 
IJ-eighths^Gf- 
eighths. 



Reduce to 3rds. 



Ji! i4off^M*= 

I O O 



= of 
3-thirds. 



III. .-. i~ 



Explanation. In taking \ of f , we must divide the numer- 
ator by 2. The denominator must be left unchanged ; for that 
is the denomination to which the given fraction is to be reduced. 
I. Reduce f to llths. 

1 . f ^rf^ll-elevenths. 

i i _r 11 y~ ?j- i O f 11-elevenths 

t-i of ii_ n _ n _^ _, 



II. 



-3 times 
S 




2i-elevenths. 

-^= 3 times 24-elevenths 
~ll =6felevenths. 



to a mixed number. 



3-thirds=l. 

29-thirds=as many times 1 as 3- 

thirds is contained in 29-thirds, 

which is 9f . 



III. 



I. Reduce f . f , | to their I,. C. Denominator. 

1. L. C. D.=12. 

2. 

3. 

4. 

II.- 




48 



FINKEI/S SOLUTION BOOK. 
I. Reduce |, |, | to their L. C. DenoMrntf-or 



fl. L. C. D.=40. 

2. !=. 

li. 3. 4=4x1*=!*. 



i- ' i, I, *=-**, H 

I. Add |, f, J. 
1. L. C. D.=24. 

2. I=. 



4. |=fxM= 
5- J=x= 

6. .'. 



I. Reduce f to a fraction whose numerator is 15. 



III. .-. f=if. 

I. Reduce f, f , ^ to equivalent fractions having 
least common numerators. 

1. L. C. N.=12. 

2. 1=||. 

A 
4. 



in. .-. 



I. Subtract f from - 
I. L. C. D.=40. 

2- 1=4*- 

3- f=fX=it- 

4- - 



TIL ,. A-t^U, 
1. Multiply J by f. 



fl- 

n.{2. 

13. 



f X|=5 times 

III. .'. X*=l: 



FRACTIONS. 



I. Divide -| by ^. 
II. J 2. |44=7 times f=V- 

U. 4-HN4 of =44=H. 



in. ... 

Analysis to the last example : 

'1. nj- is contained in 1, or J-, 7 times. 

2. f is contained in l,orf, ^ of 7 times |- times. 

3. f is contained in l, ^ of J times^-^T times. 

A- 7 is contained in -|, 5 times ^ times, or |^J times. 
Note. By inverting the divisor, we find how many times it is 
contained in 1. 

EXAMPLES. 

1. One-fifth equals how many twelfths? 

2. Reduce -|, ^, -|, and \ to fractions having a common de- 
nominator. 

1 ^ 

3. Reduce I- to a fraction whose numerator is 13. Ans. _ ! 



4. 

5. 
6. 

7. 
8. 
9. 

10. 
11. 



i 
1 

1 
13. l=what? ^4 ns. 907200. 



Reduce to a fraction whose denominator is 11. Ans. _ 

11 

Reduce ~|, -$-, f, to fractions having common numerators. 
Add , -J, |l, |, and ^. 
3 O f 8f | of 5 what? 
Multiply f by 8f . 
Multiply I of 9i of f by f of 17. 
197 r;2 

1Z . Of 






Ans. 






a 

14. !i= 



4 



50 FINKEL'S SOLUTION BOOK. 

15. (2iX2*+* of T V) X (f) 3 -H7f-3iXff)=what? 



400000 
Ans ' 407511: 



. A+H-A _ 

" 



-7H 62A 4f - ' 4fx5i-200|^ " 3 

8 



17. 2-~-2-j-2-;-2H-2-r-2-r-2-r-2-: i : | : ^ : j 

Ans. 1. 

18. Reduce f to thirds. Ans. 2f thirds. 

19. What fraction is as much larger than f as f is less than f ? 

Ans. if. 

20. What is the value in the 13th example if a heavy mark be 
drawn between ^ and -J-. Ans. If. 



21. l--=what? 



22. Subtract * of from ^ of -j- Ans. f JJJ. 

*t f 

23. What is the relation of 11 to 3? 

(1. I=iof3. 

Solution: 4 2. 11=11 times J of 3=Jf of 3=3 
( times 3. 

Conclusion: /. 11 is 3f times 3. 

24. What is the relation of 19 to 5? Ans. 3J. 

25. What is the relation of T 6 T to 24? Ans. ^ 

26. What part of 3 is 2? 

Solution: 



i 
2. 2=2 times J of 3= of 3. 

Conclusion: .-. 2 is of 3. 

27. What part of 6 is 7? Ans. J.. 

28. What part of 3 is J? Ans. T V 

29. What part of 4 is 3? Ans. -^ 

30. What part of f of f is f of T V 



EXAMPLES. 51 

ri. |oft=f 

Solution: f f f A f -|V 

( 4. ^ is A times f of f = T 7 2- of f . 
Conclusion: .'. f of T 7 ^, or ^, is T 7 ^ of f of f . 

i + i,! 

31. 24i= what? Ans. 1&. 

[Note. This is a continued fraction.] 

32. Find the number of which 75 is ^. 

33. Find the number of which 180 is f . 

34. if is f of what number? 

11. f of some number=if . 
I of that number=i of =*. 
o. I oi that number, or the number rer 
quired, =4 times T % if . 

Conclusion: .-. if is f of if. 

35. 27 is .3 of what number? Ans. 90. 

36. f of if is \ of ,f of what number? 



*' . '' ' f =* of som j nun i ber or 

4. ^ of some number=. 

5. f of that number, or that number, 



Solution : 



Conclusion: .-. of |f is J- of f of -J 5 -. 

37. A watch cost $30, and this is f of f of the cost of the 
watch and chain together. What did the chain cost. Ans. 



38. A lost f of his money and then found f as much as he 
lost and then had $120; how much money had he at first? 

39. A sum of money diminished by f of itself and $6 equals 
$12; what is the sum? Ans. 



40. If % of a ton of hay is worth $8^, how much is 10 tons 
worth? Ans. $204. 

41. What number is that if of which exceeds -fa of it by 
111? Ans. 216. 

42. What part of 2J feet is 3J inches? Ans. ^ 

43. A has $2400; f of his money plus $500 is } of B's; what 
sum has B? Ans. $1600. 



52 FINKEL'S SOLUTION BOOK. 



44. What fraction of -^V TT-^T 1 I -^T+^^T 1 

Ans. 



45. A pole stands f in the mud, -^ in the water, and the re- 
mainder, 12f feet, above water. Find the length of the pole? 

Ans. 44f feet. 

46. If 48 is ^ of some number, what is f of the same num- 
ber? Ans. 63. 

47. A can do a certain piece of work in 8 days, and B can do 
the same work in 6 days. In what time can both together do 
the work? Ans. 3^ days. 

48. The lesser of two numbers is - ' g Qa , and their differ- 

7 * f 

ence is - |-. What is the greater number? Ans. -^-jp-- 

*r 

49. What number multiplied by f of f X3f will produce |f ? 

Ans. f . 

50. What number divided by If will give a quotient of 9J? 

Ans. *$ 

51. A post stands in mud, J in water, and 21 feet above 
the water? What is its length? Ans. 36 feet. 

52. A can do a piece of work in 8 days, A and B can do it in 
5 days, and B and C in 6 days. In what time can A, B, and C 
do the work? Ans. 3 T % days. 

53. If f of 6 bushels of wheat cost $4J, how much will f of 
1 bushel cost? Ans. 80 cents. 

55. What number diminished by the difference between ^ and 
of itself leaves 1152? Ans. 2268. 

56. If a piece of gold is f pure, how many carats fine is it? 

. Ans. 15 carats. 

57. The density of the earth is 5f times that of water, and 
the sun is \ as dense as the earth. How many times denser than 
water is the sun? Ans. . 



CIRCULATING DECIMAL: 



53 



CHAPTER XI. 

CIRCULATING DECIMALS. 
I. Change .63 to a common fraction. 

.63=.636363+ etc., ad injinitum. 

.636363+etc.,=.63+.0063+.000063+etc., ad infinitum. 
13. This is a geometrical infinite decreasing series whose 
first term is .63 and ratio .63-7-.0063= T fo. The sum 

of such a series is =.63-r-(l 5-^)=!-^=^-. 

III. .-. .63= T 7 T . 

I. Reduce 1.001 to a common fraction. 

1. i.OOi=l.q0110011001100H0011+etc., ad infinitum. 

2. 1.001=1.0011. 

3. 1.00li= l+.0011+00000011 + 000000000011+etc., ad 

infinitum. 



II., 



4. .60li=rirst term. 

5- TTmnr=- 001 l-r-00000011=ratio. 

6. . 



17. .'. 

(Ray's H. A., p. 120, ex. 8.) 

III. /. 1.001=1*^. 

Remark. Since the denominator of the ratio is always ten or 
some power of ten, the numerator of the* fraction resulting from 
subtracting the ratio from 1, will have as many 9's in it as there 
are ciphers in the denominator of the ratio. By dividing the first 
term by this fraction, its numerator becomes the denominator of 
the fraction required. Hence, a circulate may be reduced to a 
common fraction by writing for the denominator of the repetend 
as many 9's as there are figures in the repetend. Thus, .63 



I. Reduce .034639 to a common fraction. 
i 034639 - 034 * - 34 ttl.- 34 X 999 +639_ 34 X 999+639 
~-TO 4 ttf =3! Tooo VWr 1000X999 ' 

__ 34 X (10001) +639 34000 34+639 34000+639 34 

9990<X) 999000 999000 

34639 34_ 34605 __ 6921 = 2807 769 
999000 ~999000~199800 66600 ~22200' 



54 FINKEL'S SOLUTION BCQK. 

In case the circulate is mixed, we have the following rule : 

1. For the numerator, subtract that part which precedes the 
repetend from the whole expression, both quantities being consid- 
ered as units. 

2. For the denominator, write as many 9*s as there are figures 
in the repetend, and annex as many ciphers as there are decimal 
fgures before each repetend. 

I. ADDITION OF CIRCULATES. 

I. Add 5.0770, .24, and 7.124943. 

(1. 5.0770 = 5.0770 = 5.07707707 etc. 
2. .24 = .242 = .24242424 etc. 
3. 7.124943= 7.124943J= 7.124943J2 etc. 
III. .-. Sum=12.44 12.4444444 etc.=12.44. 

Explanation. The first thing, in the addition and subtrac- 
tion of circulates, is to make the circulates co-originous , /. e. , to 
make them begin at the same decimal place. That is, if one be- 
gins at (say) hundredths, make them all begin at hundredths, 
providing that each circulate has hundredths repeated. It is 
best to make them all begin with the circulate whose first re- 
peated figure is farthest from the decimal point, though any 
order after that may be taken. In the above example we have 
made them all begin at hundredths. After having made them 
all begin at hundredths, the next step is to make them con- 
terminous, i.e.) to make them all end at the same place. To 
do this, we find the L. C. M. of the numbers of figures repeated 
in each circulate, then divide the L. C. M. by the number of fig- 
ures repeated in each circulate for the number of times the figures 
as a group must be repeated. Thus, the number of figures in the 
first repetend is 3; in the second, 2; and in the third, 6. 

The L. C. M. of 3, 2, and 6 is 6. 6-r-3=2. .*. 770 must 
be repeated twice. 6-j-2=3. .'. 42 must be repeated three 
times. 6--6=l. .'. 249431 must be taken once. 
I. Add .946, .248, 5.0770, 3.4884, and 7.124943. 

fl. .946 = .946 = .946666666666666 etc. 

2. .248 = .2484 = .2484-84848484848 etc. 

3. 5.0770 = 5.07707 = 5.077077077077077 etc. 
IL<U. 3.4884 = 3.488448 == 3.488448844884488 etc. 

5. 7.124943= 7.12494312 = 7.124943124943124 etc, 

6. Sum =16.88562056205620+, 
=16.885620. 

III. .-. Sum=16.885620. 




III. 
I. 

II. 
III. 

I. 



I. 



CIRCULATING DECIMALS. 



II. SUBTRACTION OF CIRCULATES. 
Subtract 190.476 from 199.6428571 

1. 199.6428571 = 199.64285714 

2. 190.476 = 19Q 47619047 



55 



I Difference = 9.1666666 = 9.16. 
.-. Difference=9.16. 
Subtract 13.637 from 104.1. 
1. 104.1 =104.14 =104.1414141 etc. 
13.637= 13.637= 13.6376376 etc. 



-3. 



Difference = 90.503776 

.-. Difference=90.503776. 

III. MULTIPLICATION OF CIRCULATES. 

Multiply .07067 by .9432. 
.07067=.070677 
.9432 = .9^?- Multiply by the fraction thus : 

.06609 .070677 

.003056 16 

.066665=product. .42406=.424062 
.7067 =.706770 

37)1.13083(3056= 

HI 003056, be- 
cause the 
fraction is 



208 
185 



Multiply 1.256784 by 6.42081. 

1.256784=1.2567842 

6.42081 = 6.420 T 9 T 

.02513568 = .025135685 1 

.5027137 = .502713702 7 

7.540705 7.540705540 7 

.001028270 = .001028276 



233 
222 

~ 



Multiply by the 
fraction thus : 

1.256784 2 



8.069583198 



11).011311Q57 
.001628276- 



56 FINKEL'S SOLUTION BOOK. 

Remark. In multiplying by any number, begin sufficiently far 
beyond the last figure of the repetend, so that if there is any to 
carry it may be added to the repetends of the partial products, 
making them complete. Thus in the above example, when mul- 
tiplying by 4, we begin at 5, the second decimal place beyond 4, the 
last figure of the repetend ; and so when we multiply 4 by 4, thtf 
first figure of the repetend in the partial product is 7. 

IV. DIVISION OF CIRCULATES. 

RULE. Change the terms to common fractions; then divide as 
4n division of fractions, and reduce the quotient to a repetend. 

I. Divide .75 by .1 

fl. .75=H= 

II. 2. .1=4. 

[3. f-f -Hr=ifx9=fi=6.8181 etc.=6.81. 

III. .-. .75-r-.i=6.8i. 

EXAMPLES. 

1. Add .87, -8, and 876. Ans. 2.644553. 

2. Add .3, .45, .45, .351, .6468, .6468, .6468, and 6468. 

Ans. 4.1766345618. 

3. Add 27.56, 5.632, 6.7, 16.356, .71, and 6.1234. 

Ans. 63.1690670868888. 

4. Add 5.16345, 8.6381, and 3.75. 

Ans. 17.55919120847374090302. 

5. From 315.87 take 78.0378. Ans. 237-838072095497. 
6. From 16.1347 take 11.0884. Ans. 5.0462. 

7. 18 is .6 of what number? Ans. 27. 

8. From ^ take ^. Ans. .1764705882352941. 

9. From 5.12345 take 2.3523456. 

-4*5.2.7711055821666927777988888599994. 

10. Multiply 87.32586 by 4.37. Ans. 381.6140338. 

11. Multiply 382.347 by .03. Ans. 13.5169533. 

12. Multiply .9625668449197860 by .75. Ans. .72. 

13. Divide 234.6 by .7. Ans. 701.714285. 

14. Divide 13.5169533 by 3.145. Ans. 4.297. 



PERCENTAGE AND ITS VARIOUS APPLICATIONS. 57 

15. Divide 2.370 by 4.923076. Ans. 481. 

16. Divide ,36 by .25. Ans. 1.4229249011857707509881. 
17- Divide, .72 by .75. Ans. .9625668449197860. 

18. 54.0678132-r-8.594=what? . Ans. 6.290. 

19. 4.956-f-.75=what? Ans. 6.6087542. 

20. 7.714285-7-.952386=what? Ans. 8.1. 



CHAPTER XII. 

I. PERCENTAGE AND ITS VARIOUS APPLICATIONS. 

1. Percentage is a method of computation in which 100 is 
taken as the basis of comparison. 

2. Per cent, is an abbreviation from the L,atin, per centum, 
per, by, and centum, a hundred. 

3. The Terms used in percentage are the Base, the Rate, 
the Percentage, and the Amount or Difference. 

4. The Base is the number on which the percentage is 
computed. 

5. The Rate is the, number of hundredths of the base 
which is to be taken. 

6. The Percentage is the result obtained by taking a cer- 
tain per cent, of the base. 

7. The Amount or Difference is the sum or difference 
of the base and percentage. 

8. The sign, %, is used instead of the words i( per cent." 
and "one-hundredths," following the number expressing the 
rate. Thus, for example, for 5 per cent., or 5 one-hundredths, 
we write 5%. 

Hence, we have the following identical expressions: 

5 per cent. =5 one-hundredths=yf^=.05^5%. In each of 
these expressions the fractional unit is T J^. The fundamental 
principle of percentage is that our computation shall be made on 
the basis of hundredths. That this principle be not violated, 
the denominator of the fraction must always be 100. Thus, 
since T I I) %= T I Q, we can take T ^ of a number instead of ^ f it 
and get the same result; but using fractions whose denomina- 
tors are numbers other than 100 to express the rate is not the 
method of percentage, but merely the method of common frac- 
tions. However, in teaching percentage the method of common 



58 FINKEL'S SOLUTION BOOK. 

fractions should also be used, as this method, because of its 
brevity, is more often used in practice. 

As an illustration, find 5% of $600. 

1. 100 one-hundredths, or {%%, or 1.00, or 100%=$600, 

2. 1 one-hundredth, or -j-fr , or .01, or 1%,=^ of 
II. ^j $600=$6, 

3. 5 one-hundredths, or T ^, or .05, or 5% 5 times 



III. /. 5% of $600=$30. 
I. What is 8% of 150 yards? 



FIRST SOLUTION. 



I. m=lW yards. 
II. 2. T^Tfo of i$r=Tiir of 150 yards=1.5 yards. 

3. y^ =:8 times 1.5 yards=12 yards. 
III. /.8% of 150 yards=:12 yards. " 

Remark. This solution is by the method of percentage purely. 



II. \ 2. 
(3. 



SECOND SOLUTION. 



1. 100%=:150 yards. 

1%=-^ of 100%=^ of 150 yards=1.5 yards. 
8%=8 times 1% 8 times 1.5 yards=12 yards. 



III. /.8% of 150 yards=12 yards. 

THIRD SOLUTION. 

Briefly, by fractions : 

8% of 150 yards^y^ of 15(0 yards=12 yards. 



CASE I. 

P tage. 

Formula. BXR=P, where B is the base, R the rate, and 
P the percentage. 



PERCENTAGE AND ITS VARIOUS APPLICATIONS. 59 

I. What is 8% of $500? 

r l. 100% $500, 
II J 2. 1 % y^ of $500 $5, and 

U. 8% 8 times $5 $40. 
III. .-. 8% of $500 $40. 

1. What is |% of 800 men? 
-1. 100% 800 men. 

1% YTFTT f 800 men 8 men, and 
^%=| times 8 men 6 men. 
III. .-. |% of 800 men 6 men. 



II./2. 
U. 



II 



I. What is 10% of 20% of $13.50? 

1.50. 

2. l%= T i<> of $13.50 $.135, and 
.3. 20% 20 times $.135 $2.70. 



(2.) 100% $2.70. 



(3.) l%=riir of $2.70 $.027, and 

(4.) 10% 10 times $.027 $.27 27 cents. 

III. .-. 10% of 20% of $13.50 27 cents. 

I. A. had $1200 ; he gave 30% to a son, 20% of the remain- 
der to his daughter, and so divided the rest among four 
brothers that each after the first had $12 less than the 
preceding. How much did the last receive? 

rl. 100% $1200, 
J2. 1% T fo of $1200 $12, and 
( ') 3. 30% 30 times $12 $360 son's share. 
U. $1200 $360 $840 remainder, 
rl. 100% $840. 

I 2. 1%= 1 ^ of $840 $8.40, and 

(2. W3. 20% 20 times $8.40 $168 daughter's share. 
j4. $840 $168 $672 amount divided among four 

brothers. 

(3.) 100% fourth brother's share; 
(4.) 100%+$12=--third brother's share. 
(5.) 100% +$24 second brother's share, and 
(6.) 100% + $ 36 = first hi other's share. 
(7.) 100%+(100%+$12)-j-(100%+$24)+(100%+ 
$36) 400%+$72 am'tthe four brothers rec'd. 
(8.) $672 amount the four brothers received. 
(90 ..400%-f$72 $672. 
( 10. ) 400 % $672 $72 $600. 
/ 1 1 \ ~\ ot i f\f 'tfinri *ki ^n 

IXJL.J -I m TlTlf <pOUU pl.c'v/. 

(12.) 100% 100 times $1.50 $150 fourth brother's 
share. 

III. .-. The last received $150. (R. H. A., p. 191,prob. 25.) 



60 FINKEL'S SOLUTION BOOK. 

1. What number increased by 20% of 3.5, diminished by 
12^% of 9-6, gives 3|? 

100%=the number. 
. 100% =3.5, 

; . 1%= T ^- of 3.5=.035, and 
.3. 20% =20 times .0*5=.7. 
TT J r l. 100% =9.6, 

(3.)<|2. 1%=TF(T of 96=.096, and 

12%=12-J times .096=1.2. 
(4.) .-. 100%+-7 1.2=3i, 
(5.) 100% .5=3.5, and 
1(6.) 100%=4, the number. 
III. .-. The number=4. (/?. H. A., p. 191, prob. 26.} 

CASE II. 

Given j the base * nd the I to find the rate per cent. 
( percentage j 



Formula. P^B=R, where B is the base, P the percentage 
and R the rate per cent. 

I. 750 men is what % of 12000 men? 

12000 men=100%, 

2. 1 man= T 2^o of 100%= T f ff % , and 

,3. 750 men=750 times -&*%*=&\%- 
III. .'. 750 men is 6^% of 12000 men. 

I. A's money is 50% more than B's; then B'sishow many % 
less than A's? 

II. 100%=B.'s money. Then, 
2. 100%+50%=150%=A.'s money. 
3. 150%=100% of itself. 
4. 1 %= T io of 100%=|% , and 
5. 50%=50 times f %=33%. 
III. .-. B.'s money is 33^% less than A.'s 

(R. H. A.,p. 192, prob. 11.) 

I. 30% of the whole of an article is how many % of f of it? 

'1. 100%=whole article. 
2. 66f %=f of 100%=f of the article. 
? =100% of itself. 



4. 1%= of 100%=H%, and 

bo 3 

5. 30%=30 times U%=45%. 



Ill .-. 30%-of the whole of an article is 45% of f of it, 

(R. H. A., p. 192, prob. 20.) 




I. 



PERCENTAGE. 



61 



If a miller takes 4 quarts for toll from every bushel he 
grinds, what % does he take for toll? 

(1. 1 bu. 32 qt. 
TT 1 2. 32 qt. 100%, 
'13. 1 t.= 



V 4. 4 qt.=4 times 3^% 12|%. 
III. .'. He takes 12|% for toll. 



CASE III. 



~ . ( tne percentage and ) . r , , , , 
Glven to find the base ' 



I the rate per cent 



d ) . 
. } to 



Formula. P-^RB, where P is the percentage, R the rate 
per cent., and B the base. 



$24 is f% of what sum? 

1. 100%=sum. 

2. |%=$24, 

3. i%=4 of $24=48, 

4. , or 1%,=8 times $8=$64, and 






III. 



I. 



5. 100% 100 times $64 $6400. 
.-. $24isf% of $6400. 

I drew 48% of my funds in bank, to pay a note of $150; 
how much had I left? 



III. 



100% amount in bank. 

48% amount drawn out. 
100% 48% 52% amount left. 

48% $150, 
!%=& of $150 $3.125, and 

52% 52 times $3.125 $162.50 amount left. 

$162.50 amount I had left. 






III. 



I pay $13 a month for board, which is 20% of my salary; 
what is my salary? 

100% my monthly salary. 

20% $13, 

l%=^r of $13 $.65, and 
100% 100 times $.65 $65, my monthly salary. 

.'. $78012 times $65 my yearly salary. 

My salary=$780. (R. H. A,, p. 194, prob. 20.) 



I. 
2. 
3. 
4. 
5. 



62 



FINKEL'S SOLUTION BOOK. 



CASE IV. 



Given 



\ the Unt 

( 



to find the base. 



, 
rate per cent. 

Formula. A-*-(\+R)=B, where A is the amount, that is, 
the base and the percentage, R the rate per cent., and B the 
base. 

III. .-. $540 is 8% greater than $500. 

I. A sold a horse for $150 and gained 25%; what did the 
horse cost? 

1. 100%=cost of horse. 

2. 25%=gain. 

3. 100%+25%=125%=selling price of horse, and 

4. $150=selling price of horse ; 

.'. 125%=$150, 

of $150=$1.20, and 



III. 



5. 
6. 
7. 100%=100 times $1.20=412Q=cost of horse. 

.-. The horse cost $120. 



I. I sold two horses for the same price, $150; on one I 
gained 25% and on the other I lost 25%; what was the 
cost of each? 

1. 100%=cost of first horse. 

2. 25%=gain. 

3. 100%+25%=125%=selling price of first horse, 
A. <4. $150=selling price of first horse; 

5. .'. 125%=$150, 

6. 1%= T ^. of $150=$1.20, and 

7. 100%=100 times $1.20=$120=cost of first horse. 
'1. 100%=cost of second horse. 

2. 25%=loss on second horse. 

3. 100% 25%=75%=selling price of 2d horse, and 
B <J4. $150=selling price of second horse; 

5. .-. 75%=$150, 

6. 1 %= T V of $150=$2, and 

7. 100%=100 times $2 $200=cost of second horse. 



TTT 
11 



J$120=cost of first horse, and 
'* 1 $200=cost of second horse. 



I. A coat cost $32; the trimmings cost 70% less, and the 
making 50% less than the cloth; what did each cost? 



PERCENTAGE. 



63 



II. 



III. 



II. 



III. 



r 1. 

2. 
3. 
4. 
5. 
6. 

8. 

9. 

10. 



100%=cost of cloth. Then 
100% 70%=30%=cost of trimmings, and 
100% 50%=50%=cost of making. 
100%+30%+50%=180%=cost of coat. 
$32=cost of coat; 



^- of $32=$. 



" 



IOO%=tlOO times $.1777=$17.77=cost of cloth. 
30%=30 times $.1777J=$5.33i=cost of trimming. 
50%=50 times $.1777J=$8.88f=cost of making. 

r$17.77J=cost of cloth, 

<[$ 5.33^=cost of trimmings, and 

^$ 8.88|=cost of making. 

(R. H. A., p. 196,prob. 12.) 



n a company of 87, the children are 37^% of the women, 
who are 444% of the men; how many of each? 



(1.) 

(2.) 



(4-) 

(5.) 
(6.) 

(V.) 

(8.) 

(9.) 

.(10.) 



100% number of men. Then 
44f%=number of women. 



3. 37^%=37|- times .44f %=16f %=number of chil- 
dren in terms of the number of men. 

10C%+44f%-f-16f% = 161^-%= number in the 
company, 

87 number in the company ; 



l%=i of 87=.54, 

100%-=100 times .54==54=number of men, 
444%==44 times .54=24=number of women, and 
16|%=16| times .54=19=number of children. 



54=number of men, 
24=number of women, and 
19=number of children. 



H. A., p. 197,prob. 20.) 



I. Our stock decreased 33^%, and again 20%; then it rose 
20%, and again 33^-%; we have thus lost $66; what 
was the stock at first? 



64 FINKEL'S SOLUTION BOOK. 



100%=original stock. 

33i%=decrease. 

100% 33%=66f %=stock after first decrease. 

100%=66f%, 

!%= T i T of 66J%=f%, and 
20% 20 times f %=13%=second decrease. 
66f % 13i%=53%=stock after second decrease. 



20%=20 times .53%=10f %first increase. 
II.<| U. 53^%+lOf %=64%=stock after first increase 

2. 1% =T ^ of 64%=.64%, and 
I 33%=33| times .64%=21%=second increase. 
[. 64%-f-21%=85i|-%=stock after second increase. 

100% 85^%=14f %=whole loss; 

$66=whole loss; 

... 14| %= $66; 

HO.) I%=:rr5 of $66 $4.50, and 
\ / / 14 

(11.) 100%=100 times $4.50=$450=original stock. 
III. .-. $450=original stock. 

I. A brewery is worth 4% less than a tannery, and the tan- 
nery 16% more than the boat; the owner of the boat 
has traded it for 75% of the brewery, losing thus $103 ; 
w'nat is the tannery worth ? 

FIRST SOLUTION. 

(1.) 100 %=value of the tannery. Then 
(2.) 100% 4%=96%=value of the brewery. 

'1. 100%=value of the boat. Then [the boat. 

2. I00%+16%=116%=value of tannery in terms of 

3. 116%=100%, the value of tannery from step (1), 

4. l%=nhr of 100%=ff % , and 

5. 100%=100 times jfr%=86ff%=v&lue of the boat 

in terms of the tannery. . 
\. 100%=96%, 
llA (A \J2- 1%= T *irof96%=.96%, and 

75%=75 times .96%=72%=what the owner 

of the boat received for it. 

' 862%% 72% 14^V%=what the owner of the 
boat lost in the trade. 
(6.) $103=what he lost; 
(7.) ... 14^g-%=:$103, 

(8.) l^^L-of $103 $7.25, and 

(9.) 100%=100 times $7.25=$725=value of tannery. 
III. .'. $725=value of the tannery. (R.H.A.,p.W7,prob.2S.) 



(3.) 




PERCENTAGE. 



65 



Remark. The value of the brewery and boat being ex- 
pressed vi terms of the tannery, 75% of the brewery is also ex- 
pressed in terms of the tannery; hence, it is plain that the owner 
of the boat has traded 86^ 6 9 % for 72% of the same value, losing 

86/^% 72%, or 14^%. 

SECOND SOLUTION. 

(1.) 100% value of the boat. Then 

(2.) 100%+16% 116% value of the tannery. 

1(1. 100% 116%, 
(3.K2. 1%=^ of 116% 1.16%, and 
l3. 4% 4 times 1.16% 4.64%. 
(4.) 116% 4.64% 111.36% the value of brewery in 
terms of the boat, 
rl. 100% 111.36%, 
\\\ ,,. J2. 1%^ of 111.36% 1.1136%, and 

I') 3. 75% 75 times 1.1136% 83.52% what the 
owner of the boat received for it. 
(6.) .-. 100% 83.52% 16.48% what he lost in the 
trade. 
(7.) $103 what he lost. 
(8.) .', 16.48% $103, 
(9.) l%=nr?frir of $103 $6.25, and 
.(10.) 116% 116 times $6.25 $725 value of tannery. 
III. .-. $725 value of the tannery. 

THIRD SOLUTION. 

100% value of brewery. 

100% value of tannery. Then 

100% 4% 96% value .of the tannery. 

.-. 96%. 100%, the value of brewery in step (1), 

l%=-fa of 100% 1.041%, and 

100% 100 times 1.04^% 104% value the tan- 
nery in terms of the brewery. 

100% value of boat. Then 

100%+16% 116% value of the tannery in 
terms of the boat. 

.-. 116% 104^%, the value of the tannery in step 
5 of (2), 
1% rle of 104^% .89^1% , and 

100% 100 times .89if% 89 jf% value of the 
boat in terms of the tannery, and consequently 
in terms of the brewery. 

.* 89-H1 % 75% 14iff % what the owner of 
the boat lost in the trade. 

$103 what the owner of the boat lost ; 



(2.) 



(3.) 



(4.) 

(5.) 
(6.) 

(7.) i% = _L^ of $103=16.96, and 

I /\ M 

I (8.) 104i% 104^ times $6.96 $725=value of tannery. 



66 FINKEL'S SOLUTION BOOK. 

III. .-. $725=value of of the tannery. 

Remark. In step 5 of (3), we have the value of the 'boat in 
terms of the tannery ; but the value of the tannery is in terms of 
the brewery: hence, the value of the boat is also in terms of the 
brewery. The owner of the boat, therefore, traded 89|ff % for 
75% of the same value. 

MISCELLANEOUS PROBLEMS. 

I. A man sold a horse for $175, which was 12-^-% less than 
the horse cost; what did the horse cost? 

1. 100% cost of horse. 

2. 12|% loss. 

3. 100% 12|% 87i% selling price. 
, 4. $175 selling price. 

IL< >5. .-. 87i% $175, 

6. 1^=_L O f $175 $2, and 

7. 100% 100 times $2 $200, 

III. .-. $200 cost of the horse. (R. 3d p., p. 204, P r t>- 5.) 

I. A miller takes for toll 6 quarts from every 5 bushels of 
wheat ground; what % does he take for toll.? 

1. 1 bu. 32 qt. 

2. 5bu. 5 times 32 qt. 160 qt. 



II. 



4. 160 qt. 100% 



c, 



4. 1 qt. T ^ of 100% f%, and 

5. 6qt.6 times |%3f%. 

III. " .-. He takes 3f % for toll. (R. 3d p., p. 204, prob. H-) 

I. A farmer owning 45% of a tract of land, sold 540 acres, 
which was 60% of what he owned; how many acres were 
there in the tract? 
(1.) 100% number of acres in the tract. 

1. 100% numbers of acres the farmer owned. 

2. 60%=number of acres the farmer sold. 

3. 540 acreswhat he sold. 
(2.)J4. .-. 60% 540 acres, 

5. 1%=^. of 540 acres=9 acres, and 
ll.{ 6. 100%=100 times 9 acres 900 acres=what he 

owned. 

(3.) 45% what he owned. 

(4.) .-.45% 900 acres, 
(5.) 1%=^ of 900 acres 20 acres, and 

(6.) 100% 100 times 20 acres 2000=number of acres 

in the tract. 
III. .-. The tract contained 2000 acres. 

(R. 3d p., p. 204,prob. 12.) 



MISCELLANEOUS PROBLEMS. 



67 



I. 



!!.< 



A, wishing to sell a cow and a horse to B, asked 150% 
more for the horse than for the cow; he then reduced 
the price of the cow 25%, and the horse 33%, at which 
price B took them, paying $290; what was 'the price of 

each ? 



1. 

2- 
(3. 

(4. 



100% asking price of the cow. Then 
100%-H50%=250%=asking price of the horse. 
100% 25%=75%=selling price of cow. 
100%=250%, 

l%=nnr of 250%=2.50%, and 
33%=33l times 2.5%=83%=reduction on the 

asking price of the horse. 

250% 83i%=166f%=selling price of the horse. 
75%+166f%=241f%= selling price of both. 
$2QO=selling price of both. 
/. 241f%=$290, 

f 290 == 1 - 20 and 



) 75% 75 times$ 1.20=$90=selling price of the 

cow. 
) 166S%=166f times $1.20=$200=selling price of 

the horse. 



f$ 90=selling price of the cow, and 
''|$200=selling price of the horse. 

(Brooks' H. A., p. 2J.3, prob. 18.) 




I. A mechanic contracts to supply dressed stone for a church 
for $87560, if the rough stone cost him 18 cents a cubic 
foot; but if he can get it for 16 cents a cubic foot, he 
will deduct 5% from his bill; required the number of 
cubic teet and the charge for dressing the stone. 

100%=$87560. 

1%= T ^ of $87560=$875.60, and 

5% = 5 times $875.60=$4378=the deduction. 

18/ 16/=2/ the deduction per cubic foot. 

... $4378=the deduction of 4378-7-.02, or 218900 cubic 
feet. Then 

$87560=cost of 218900 cubic feet. 

$.40=$87560-r-218900=:cost of one cubic feet. 

.-. $.40 $. 18=$. 22=cost .of dressing per cubic foot. 



Ill 



r218900=number of cubic feet, and 
' '122 cents=cosc of dressing per cuoic foot. 

(Brooks' H. A., p. 



, prob. 21.) 



68 FINKEL'S SOLUTION BOOK. 



EXAMPLES. 

1. A merchant, having $1728 in the Union Bank, wishes to 
withdraw 15%; how much will remain? Ans. $1468.80. 

2. A Colonel whose regiment consisted of 900 men, lost 8% 
of them in battle, and 50% of the remainder by sickness; how 
many had he left? Ans. 414 men. 

3. What % of $150 is 25% of $36? Ans. 6%. 

4. What % of I of f off is -J? Ans. Bl%. 

5. If a man owning 45% of a mill, should sell 33-j-% of his 
share for $450 ; what would be the value of the mill ? 

Ans. $3000. 

6. A. expends in a week $24, which exceeds by 33^% his 
earnings in the same time. What were his earnings? Ans. $18. 

7. Bought a carriage for $123.06, which was 16% less than I 
paid for a horse; what did I pay for the horse? Ans. $146.50. 

8. Bought a horse, buggy, and harness for $500. The horse 
cost 37-^% less than the buggy, and the harness cost 70% less 
than the horse ; what was the price of each? 

Ans. buggy $275ff , horse $172^f, and harness $51f. 

9. I have 20 yards of yard- wide cloth, which will shrink on 
sponging 4% in length and 5% in width; how much less than 
20 square yards will there be after sponging? Ans. l|~f yards. 

10. A. found $5; what was his gain %? Ans. oo. 

11. The population of a city whose gain of inhabitants in 5 
years has been 25%, is 87500 ; what was it 5 years ago? 

Ans. 70000. 

12. The square root of 2 is what % of the square root of 3 ? 

Ans. /~ 



13. A laborer had his wages twice reduced 10%; what did 
he receive before the reduction, if he now receives $2.02-^ per 
day? Ans. $2.50. 

14. The cube root of 2985984 is what % of the square root of 
the same number? Ans. 



15. A man sold two horses for the same price $210 ; on one he 
gained 25%, and on the other he lost 25%; how much did he 
gain, supposing the second horse cost him f as much as the first? 

Ans. $10. 



COMMISSION. 69 

16. A merchant sold goods at 20% gain, but had it cost him 
$49 more he would have lost 15% by selling at the same price; 
what did the goods cost him? Ans. $119. 

17. If an article had cost 20% more, the gain would have 
been 25% less; what was the gain % ? Ans. 50%. 



1 ^ 



II. COMMISSION. 



1. Commission is the percentage paid to an agent for the 
transaction of business. It is computed on the actual amount of 
the sale. 

2. An Agent, Factor, or Commission Merchant, 

is a person who transacts business for another. 

3. The Net Proceeds is the sum left after the commission 
and charges have been deducted from the amount of the sales or 
collections. 

4. The Entire Cost is the sum obtained by adding the 
commission and charges to the amount of a purchase. 

I. An agent received $210 with which to buy goods ; after 
deducting his commission of 5%, what sum must he 
expend ? 

1. 100%=what he must expend. 

2. 5%=his commission. 

3. 100%+5%=105%=what he receives. 
II. S 4. $210=what he receives. 

6. 1 %=rhr of $210=$2, and 

7. 100%=1C9 times $2=$200=what he expends. 

III. .-. $200=what he must expend. 

(R. 3d p., p. 207,prob. 4.) 

Note. Since the agent's commission is in the $210, we must 
not take 5% of $210; for we w*ould be computing commission on 
his commission. Thus, 5% of ($200+$10)=$10+$.50. This is 
$.50 to much 

I, An agent sold my corn, and after reserving his com- 
mission, invested the proceeds in corn at the same price; 
his commission, buying and selling was 3%, and his 
whole charge $12; for what was the corn first sold? 



70 



FINKEL'S SOLUTION BOOK. 



II. 



in. 

i. 



a 

3.) 



(4.)- 



100% cost of the corn. 
3% the commission. 
100% 3% 97% net proceeds, which he invested 

in corn. 

1. 100% cost of second lot of corn, 
commission. 



entire cost of second lot of- 



=co&t of second 



= of $12 $2.06, and 



3. 100%+3% 103%: 

corn. 

4. 97% entire cost of second lot of corn. 

5. .-. 103% 97%, 

6- 1 %=^ of 97% jfa% , and 

7. 100% 100 times $*%=&fo% 

lot of corn in terms of the first. 

8. 3% 3 times T y^% 2 T 8 -^% com mission on 
second lot. 

(5. 3%+2 T 8 -0 5 3-%5 T 8 TF \% whole commission. 

(6. $12 whole commission. 

(7- 

(8.) 

(9. ) 100% 100 times $2 06 $206 cost of first lot of corn 
. $206 cost of first lot of corn. (R.H. A., p. 219,prob. 10.) 

Sold cotton on commission, at 5%; invested the net pro- 
ceeds in sugar, commission, 2%; my whole commission 
was $210 ; what was the value of the cotton and sugar? 
r (1. 100% cost of cotton. 

(2. 5% commission. 

(3. 100% 5% 95% net 

1. 100%' cost of sugar. 

2. 2 % commission. 

3. 102% entire cost of sugar. 

4. 95% entire cost of sugar. 

5. .-. 102% 95%, 

6. 1%=^ of 95% T 9 Ti 5 Y%, and 

7. 100% 100 times T 9 ^% 93-^% cost of sugar in 

terms of cotton. 

8. 2% 2 timesY^% 1|4% commission on the 
sugar. 

(4.) 5%4-l|T% 6|4 % whole commission. 

(5.) $210=\vhole commission. 
(6.) .-. 6|{%=$210, 

(7.) 1 % =J 4 of $210=$30.60, and 

(8.) 
(9.) 

TIT J$3060 cost of cotton, and 
'''^$2850 cost of sugar. 



[vested in sugar, 
proceeds, which he in- 



100% 100 times $30.60 $3060 cost of cotton. 

93-^ times $30.60 $2850 cost of sugar. 



. H. A., p. 219,prob. 6.) 




COMMISSION. 



71 



II. 



III 



III 
I 



A lawyer received $11.25 for collecting a debt ; his com- 
mission being 5%; what was the amountof the debt? 

1. 100%=amount of the debt. 

2. 5%=commission. 
4. $11.25=commission. 

4. .-. 5%=$11.25. 

5. l%=i of $11.25=$2.25, and 

6. 100% 100 times $2.25=$225=an)outtt O f the debt. 
.'. $225=amount of debt. 

(7?, 3d p., p. 207, prob. 6.) 

Charge $52.50 for collecting a debt of $525; what was the 

rate of commission? 
1. $525100% 
!!.<! 2. $1=^ of 100%^-%, and. 

3. $52.50=52.5 times ^-%=10%=rate of commission. 
.'. 10%=rate of commission. 

My agent sold my flour at 4% commission; increasing the 
proceeds by $4.20, I ordered, the purchase of wheat at 
2% commission; after which, wheat declining 3%, 
my whole loss was $5 ; what was the flour worth? 

100%cost of flour. 

4%=commission on flour. 
100% 4%=96%=net proceeds. 

1. 100%=cost of wheat. 

2. 2%=commission on wheat. 

3. 100%+2%=102%=entire cost of wheat. 

4. 96%+$4.20=entire cost of wheat. 

5. .-. 102 %=96% +$4.20, 

6. i%_ T ^_ f (96%+$4.20)= .94 T 2 T %+$.0411|f, 

7. 100%=100 times (.94 T 2 T %+$.0411|f )=94&%+ 

$4.11|f=cost of wheat. 

8. 2%=2 times (.94 T 2 7 %+$.0411i-f-)=ll|-%+$.08 T V 
II ^ I commission on wheat. 



[il 

(3.) 



(4.) 




5=3^- times 
loss on wheat. 

4 % + Hf % + $-08 T 4 T + 3-gV% + $ 
$.21ff=whole loss. 
$5=whole loss. 

9J T %=^$5 $.21|f $4-78^ 2 T . 
1%=J- of $4. 1 78^ r =$.53, and 



III. 



(6.) 

(7-) 
(8.) 
(9.) 

(10.) 

(11.) 100%=^100 times $.53=$53. 
.-. $53=cost of flour. ( /?. H. A., p. 219, prob. 



72 FINKEL'S SOLUTION BOOK. 



EXAMPLES. 

1. A broker in New York exchanged $4056 on Canal Bank, 
Portland, at -f- %; what did he receive for his trouble? 

Ans. $26.35. 

2. A sold on commission for B 230 yards of cloth at $1.25 per 
yard, for which he received a commission of 3^%; what was his 
commission and what sum did he remit? 

Ans. Commission $10.06^, and Remittance $277.43}. 

3. A sold a lot of books on commission of 20%, and remitted 
$160; for what were the books sold? Ans. $200. 

4. A lawyer charged $80 for collecting $200; what was his 
rate of commission? Ans. 40% 

5. I sent my agent $1364.76 to be invested in pork at $6 per 
bbl. after deducting his commission of 2%; how many barrels of 
pork did he buy? Ans. 223 bbl. 

6. How much money must I send my agent, so that he may 
purchase 250 bbl. of flour for me at $6.25 per bbl., if I pay bim 
2%% commission? Ans. $1601.5625. 

7. If an agent's commission was $200, and his rate of com- 
mission 5% ; what amount did he invest? Ans. $4000 

8. My agent sold cattle at 10% commission, and after I in- 
creased the proceeds by $18, I ordered him to buy hogs at 20% 
commission. The hogs had declined 6f%, when he sold them at 
14f% commission. I lost in all $86; whatdidthe cattle sell for? 

Ans. $200. 

9. An agent sells flour on commission of 2%, and purchases 
goods on true commission of 3%. If he had received 3% for 
selling and 2% for buying, his whole commission would have 
been $5 more. Find the value of the goods bought. 

Ans. 

III. TRADE DISCOUNT. 

1. Trade Discount is the discount allowed in the pur- 
chas and sale of merchandise. 

2. A List, or Hegular JPrice, is an established price, as- 
sumed by the seller as a basis upon which to calculate discount. 

3. A Ifet Price is a fixed price from which no discount 
is allowed. 

4. The Discount is the deduction from the list, or regu- 
.lar price. 



TRADE DISCOUNT. 



73 



II. 



III. 



Sold 20 doz. feather dusters, giving the purchaser a dis- 
count of 10, 10 and 10% off, his discounts amounting to 
$325.20; how much was my price per dozen? 

(1.) 100% whosesale price. 

(.2.1 10% of 100% 10% first discount. 

(3.) 100% 10% 90% first net proceeds. 

100% 90%, 

1% T io of 90% &%, and 
10% 10 times T 9 ^%z=9% second discount. 
90^ 9 % 81% second net proceeds, 
rl. 100% 81%, 

\3. 10% 10times ^%==8.i%==third discount. 
(6.) 10%+9%+8.1% 27.1% sum of discounts. 
(7.) $325.20 sum of discounts. 
(8.) .-. 27.1% $325.20, 
(9.) 1% ^ T of $325.20 $12, and 
(10.) 100% 100 times $12 $1200 wholesale price 

of 20 dozen. 
(11.) $60 $1200-5-20= wholesale price of 1 dozen. 

.. $60 wholesale price per dozen. 

(.ff. 3d p., p. 209, prob. 5.) 

Bought 100 dozen stay bindings at 60 cents per dozen for 
40, 10, and 7^% off; what did I pay for them? 



(I-) 
(2.) 



(3.)- 



(4.)' 



(5.). 



60/= list price of 1 dozen. 
$60100 times $.60 list price of 100 dozen. 

1. 100% $60, 

2. 1 % y^- of$60 $.60, and 

3. 40% 40 times $.60 $24=first discount. 

4. $60 $24 $36 first net proceeds. 

1. 100% $36, 

2. 1 % y-^j- of $36 $.36, and 

3. 10% 10 times $.36 $3.60=second discount. 
$36 $360 $32.40 second net proceeds. 

100% $32.40, 

1%=.^. of $32.40=$.324, and 
7|% 7| times $.324 $2.43 third discount. 



2. 
3. 
,4. $32.40 $2.43 $29.97 cost. 



I paid $29.97. 



3d p., p. 209, prob. 6.) 




I. A retail dealer buys a case of slates containing 10 dozen 
for $50 list, and gets 50, 10, and 10% off; paying for 
them in the usual time, he gets an additional 2%; what 
did he pay per dozen for the slates? 



74 FTNKEL'S SOLUTION BOOK. 

100% $50. 

2. 1 % Ho of $50 $.50. 

3. 50% 50 times $50 $25 first discount. 
A. $50 $25 $25 first net proceeds. 

100% $25. 

l^= T ^ r of$25=$.25. 

10%=10 times $.25 $2.50 second discount 
._ $25 $2.50 $22.50 second net proceeds, 
rl. 100% $22.50. 
1 2. 1%TTnr of $22.50=4.225. 

*' n s 10% 10 times $.225=$2.25=third discount. 
$22.50 $2.25 $20.25=third net proceeds. 
100%=-$20.25. 

1%=^ of $20.25=$.2025. 
2%=2 times $.2025=$.405=fourth discount. 
$20.25 $.405=$19.845=cost of 10 dozen slates. 
$1.9845=$19.845-f-10=:cost of 1 dozen slates. 

III. .'. $1.9845=cost of 1 dozen slates. 

(R. 3d p., p. 209,prob. 9.) 

I. Sold a case of hats containing 3 dozen, on which I had re- 
ceived a discount of 10% and made a profit of 12^% or 
37-JX on each hat ; what was the wholesale merchant's 
price per case? 

(1.) 37!/==profit on one hat. 

(2.) $13.5036 times $.37|profit on 3 dozen hats. 

(3.) 100% wholesale merchant's price per case. 

(4.) 10%=discount. 

(5.) 100% 10% 90%my cost. 



(6.) 12. 1%=^ of 90% .9%. 

]3. 12|% 12| times .9% lli%=profit in terms of 
wholesale price. 

(7.) .'. lli% $13.50'. 

(8.) l%==*iTT of $13.50 $1.20. 



(9.) 100% 100 times $1.20=$120=wholesale- mer- 
chant's price per case. 

HI. .'. $120=wholesale merchant's'price per case. 

(7?. 3d p., p. 212,prob. 4.) 

I. A bookseller purchased books from the publishers at 20% 
off the list ; if he retail them at the list what will be 
his per cent, of profit? 



TRADE DISCOUNT. 



II. 



1. 100%=list price. 

2. 20%=discount. 

3. 100% 20%=80%=cost. 

4. 100%=bookseller's selling price, because he sold them 

at the list price. 

5. .-. 100% 80%=20%=gain. 

6. 80%=100% of itself. 

7. 1%=-^ of 100% li%, and 

8. 20%=20 times l^%=25%=his gain %. 



(R. 3d p., p. 211,prob. 1.) 



III. .-. 25%=his % of profit. 



Note. Observe that since his cost is 80%, and his gain 
we wish to know what % 20% is of 80%. It will become evi- 
dent if we suppose the list price to be (say) $400, and then pro- 
ceed to find the % of gain as in the above solution. 

Bought 50 gross of rubber buttons for 25, 10, and 5% off ; 
disposed of the lot for $35.91, at a profit of 12% ; what 
was the list price of the buttons per gross? 

(1.) 100%=Hst price. 
(2.) 25% of !00%=25%=first discount. 

100% 25%=75%=first net proceeds. 
100%=75%, 

l%= y i of 75%=|%, and 
10%=10 times j%=7%. 
75% 7i%=67|%= second net proceeds 



ir 



1. 

2. 

3. 

,4. 
1. 

2. 1%= T ^ of 67i%=.67%, and 

3. 5%=5 times .67i%=3.375%=third discount 

4. 67t% 3.375%=64.125%=cost. 

1. 100%=64.125%, 

2. !%=.64125%,and 

3. 12%=12 times ,64125 %=7.695%=gain. 

4. .-. 64.125%-f 7.695%=71.82%=selling price. 
$35.91=selling price. 

... 71.82%=$35.91, 

1%=^^ of $35.91=$.50, and 

100% =100 'times $.50=$50=list price of 50 gross, 

$1.00=$50-;-50=list price of one gross. 

til. .'. $1.00=list price of one gross. 

(R. 3d p., p. 212,prob. 10.) 

I A dealer in notions buys 60 gross shoestrings at 70/ per 
gross, list, 50,, 10, and 5% off; if he sell them at 2d 
10, and 5% off list, what will be his profit? 



76 



II. 



FINKEL'S SOLUTION BOOK. 

70/=list price of one gross. 

$42=60 times $.70=list price of 60 gross. 

100%=$42. 



50 times $.42=$21s=first discount. 
$42 $21=$21=first net proceeds. 



10$F==10 times $.21=$2.10=second discount. 
$21 $2.10=$18.90= second net proceeds. 

100%=$18.90. 



(7.) 



5%=$.945=third discount 
$18.90 $.945=$! 7.955=cost 
100%=$42 

1 %= T Tr of $42=$.42. [count. 

20%=20 times $.42=$8.40=first conditional dis- 
4. $42 $8 40=$33.60=first conditional net proceeds. 

1. 100%=$33.60. 

2. 1 %=^ of $33.60=$.336. [discount 

3. 10%=10 times $.336=$3.36=second conditional 

4- $33.60 $3.36=$30.24= second conditional net 
proceeds. 

1. 100%=$30.24. 

2. 1 %= r f (r of $30.24=$.3024. [discount 

3. 5%=5 times $.3024=$1.512=third conditional 

.24 $1.512=$28.728=selling price. 
$28.728 $17.955=$10.773=his profit. 



III. .'. $10.773=his profit 



(R. 3d p., p. 212, prob. 9.) 



EXAMPLES. 

1. Bought a case of slates containing 12 doz. for $80 list, and 
got 45, 10, and 10% off; getting an additional 2% off for prompt 
payment, what did I pay per dozen for the slates? 

Ans. $2.9106. 

2. Bought a case of hats containing 4 doz., on which I re- 
ceived a discount of 40. 20, 10, 5, and 2% off. If I sell them at 
$4 a piece making a profit of 20%, what is the wholesale mer- 
chant's price per case? Ans. $399|Jf^. 

3. If I receive a discount of 20, 10, and 5% off, and sell at a 
discount of 10, 5, and 2^% off; what is my % of gain? 

Ans. 21|% . 

4. A bill of goods amounted to $2400; 20% off being 
allowed, what was paid for the goods? Ans. $1920. 

5. - Bought goods at 25, 20, 15, and 10% off. If the sum of 
my discounts amounted to $162.30, what was the list price of the 
goods? ' Ans. $300 




PROFIT AND LOSS. 77 



IV. PROFIT AND LOSS. 

1. Profit and Loss are terms which denote the gain or loss 
in business transactions. 



2. 
3. 

I. 



III. 



I. 



II. 



Profit is the excess of the selling price above the cost. 
LOSS is the excess of the cost above the selling price. 

A merchant reduced the price of a certain piece of cloth 
5 cents per yard, and thereby reduced his profit on the 
cloth from 10% to "8%; what was the cost of the cloth 
per yard? 
1. 100%=cost of cloth per yard. 

10%=his profit before reduction. 

8%=his profit after reduction. 
10% 8%=2%=his reduction. 
5/ reduction. 



=2 and 



100% =100 times 2^/=$2.50=cost per yard. 
.-. $2.50=cost of cloth per yard. 

(R< 3d p., p. 211,prob. 13.) 

A dealer sold two horses for $150 each; on one he gained 
25% and on the other he lost 25%; how much did he 
lose in the transaction ? 

1.) 100%=cost of the first horse; 

2.) 25%=gain. 

3.) 100%+25%=125%=selling price of first horse. 

4.) $150 selling price. 
(5.) .'. 125%=$150, 
(6.) l%= T iir of $150=11.20, and 

(7.) 100%=100 times $1.20=$120=cost of first horse. 
(8.) $150 $120=$30=gain on first horse. 

1. 100%=cost of second horse. 

2. 25%=loss. 

3. 100% 25%=75%=selling price of second horse. 



(9.) 



4. $150=selling price. 



III. 



I. 



5. .-. 75%=$150, 

6. 1%= T V of $150=$2, and 

7. 100%=400 times $2=$200=cost of second horse. 
$200 $150=$50=loss on second horse. 

$50 $30=$20=loss in the transaction. 

He lost $20 in the transaction. 

( R. 3d p. , p. 211, prod. 12. ) 

A speculator in real estate sold a house and lot for $12000, 
which sale afford him a profit of 33^% on the cost ; he 



(11.) 



78 



FINKEL'S SOLUTION BOOK. 



then invested the $12000 in city lots, which he was 
obliged to sell at a loss of 33ij%; how much did he lose 
by the two transactions? 



II. 



(!) 
(2.) 
(3.) 

(4.) 
(5.) 

(6.) 



100% cost of the house and lot. 

33^% gain. [lot. 

100%+33%=133i%=selling price of house and 

$12000 selling price of the house and lot. 

.-. 133i%=$12000. 



(7.) 
(8.) 

(9.) 
(10.) 



=^1 of $12000=190. 
100$ 



[lot. 



III. 
I. 



II. 



100% 100 times $90 $9000 cost of house and 
$12000 $9000 $3000 gain on house and lot. 
100% $12000. 

1% -^ of $12000 $120. 

33^% 33J times $120 $4000 loss on city lots. 
$4000 $3000 $1000 loss by the two transac- 
tions. 
.\ $1000 his loss by the two transactions. 

(./?. 3d p., p. 211, prob. 15. ) 

A dealer sold two horses for the same price; on one he 
gained 20%, and on the other he lost 20%; his whole 
loss was $25 ; what did each horse cost? 
(1.) 100% selling price of each horse, 
'l. 100% cost of first horse. 

20% gain on the first horse. 
100%+20% 120% selling price of first horse. 
... 120% 100%, from (1), 

1% T -b of 100%f%, and 
100% 100 times |%=83^% cost of first horse 

in terms of the selling price. 
100% 83i% 16f% gain on first horse. 

100% cost of the second horse. 

20% loss on second horse. 

100% 20% 80% spelling price of second horse. 
.-.80% 100%, from (1), 

1% V of 100% li%, and 
100% 100 times 1|%=125% cost of second 

horse in terms of the selling price. 
125% 100% 25% loss on the second horse. 
25% 16f % 8J% whole loss. 
$25 whole loss. 



(7.) 

(8.) 

(9.) 

((10.) 





1. 




2. 




3. 




4. 


(z. )< 


5. 




6. 




7. 




i. 




2. 




3! 


(3.)< 


4. 

5. 




6. 




7. 


(4.) 


(5.) 


(6.) 



l%:=_of$25=$3, and 

o^ [horse. 

100%=100 times $3=$300=selling price of each 
83%=83 times $3=$250= cost of first horse. 
125%=125 times $3=$375=cost of'second horse. 



PROFIT AND LOSS. 



79 



. r$250=cost of the first horse, and 
| $375=cost of second horse. 

I. What % is lost if f of cost equals f of selling price ? 

1. f of selling price=|- of cost. 

2. of selling price= of f of cost=f of cost. 

3. J- of selling price=4 times -J of cost=f of cost. 

4. |=cost. 

5! !=selling price. 

B. J=fr of 100%=1H%, loss. 
III. .-. Loss=lli%. 



II. 



I. 



II. 



Paid $125 for a horse, and traded him for another, giving 
60% additional money. For the second horse I received 
a third and $25. I then sold the third horse for $150 ; 
what was my % of profit or loss? 



(2.) 
(3.) 

(4.) 



(7.) 
(8. 



100%=$125, 

1%==^ of $125=$1.25, and 
60%=60 times $1.25=$75 = additional money 
paid for the second horse. 
$125+$75=$200=cost of second horse. 
$150=selling price of the third horse. 
$150+$25=$175=selling price of second horse. 
$200 $175=$25=loss in the transaction. 
$200=100%, 

$l=innr of 100%=!%, ana 
$25=25 times |%=12|%=my loss. 



III. .-. My loss is 12|%. 



(./?. H. A., p. 201,prob. 4.) 



I. If [ buy at $4 arid sell at $1, how many % do I lose? 

'1. $4=cost 

2. $l=selling price. 

3. $4 $l=$3=loss. 

4. $4=100%. 

5. $l=i of 100%=25%. 

6. $3=3 times 25%=75%=loss. 

III. .-. 75%=loss. 

I. A and B each lost $5, which was 2% of A's and 
B's money ; which had the most, and how much? 



of 



80 



FINKEL'S SOLUTION BOOK. 



II. 



III. 



(1.) 100%=A's money. 

(2.) 2J% whathe lost. 

(3.) $5 what he lost. 

(4.) .-. 2% $5, 

(5.) 1 %=^7 of $5 $1.80, and 

(6.) 100% 100 times $1.80 $180=A's money. 

1. 100% B's money. 

2. 3% what he lost. 

3. $5 what he lost. 

5. 1 %=4- of $5=$1.50, and 

3-$ 

6. 100% 100 times $1.50=$150=B's money. 

(8.) $180 $150 $30 excess of A's money over B's. 
.% A had $30 more than B. (R. H. A., p. 203, prob, 5.) 



I. Mr. A bought a horse and carriage, paying twice as much 
for the horse as for the carriage. He afterward sold the 
horse for 25% more than he gave for it, and the carriage 
for 20% less than he gave for it, receiving $577-50; what 
was the cost of each? 



I. 



100% cost of the carriage. 
200%=cost of the horse. 

20%=loss on the carriage. 

100% 20%=80%=selling price of the carriage. 
100% 200%, 

l%=TTnr of 200%=2%, and 

25% =25 times 2%=50%= gain on the horse. 
200%+50%=250%=selling price of the horse. 

80%+250%=330%=selling price of both. 
$577. 50=selling price of both. 
.-. 330%=$577.50, 

l%=irb of $577.50 $1.75, and 
100%=1 00 times $1.75=$175=cost of carnage. 
200%=200 times $1.75 $350=cost of the horse. 

> J$175=cost of the carriage, and 
' |$350 cost of the horse. 

(Milne's prac., p. 259, prob. 19. ) 

Mr. A. sold a horse for $198, which was 10% less than he 
asked for him, and his asking price was 10% more than 
the horse cost him. What did the horse cost him? 



II. 



II. 



PROFIT AND LOSS. 81 

(1.) 100% cost of the horse. 

(2.) 100% + 10% 110% asking price. 

rl. 100% 110%, 
(8..K2. l%= T io of H0% 1 T V%, and [asking price. 

13. 10% 10 times l T 1 7r %=ll% = reduction from 

(4.) 110% 11% 99%=selling price. 

(5.) $198 selling price. 

(6.) .-. 99% $198, 

(7.) 1% gV of $198 $2, and ' 

(8.) 100% 100 times $2 $200 cost of the horse. 

III. .-. $200 cost of horse. (Milne's prac., p. 259, prob. 23,) 

I. What must be asked for apples which cost me $3 per bbl n 
that I may reduce my asking price 20% and still gain 
20% on the cost? 

(1.) 100% $3. 
(2.) l%=rfo of $3 $.03, and 

(3.) 20% 20 times $.03 $.60 gain. 

(4.) $3.00+$.60 $3.60 selling price. 
'1. 100% asking price. 

2. 20% reduction. 

3. 100% 20% 80% selling price. 

4. $3.60 selling price. 

5. .-. 80% $3.60, 

6- l%=8i> of $3-60 $.045, and 

7. 100% 100 times $.045 $4.50 asking price. 

ill. .'. $4.50 asking price. (Milne's prac., p. 261, prob.38.) 



(5.) 



I. A merchant sold a quantity of goods at a gain of 

If, however, he had purchased them for $60 less than 
he did, his gain would have been 25%. What did the 
goods cost him ? 

100% actual cost of goods. 

20% gain. 

100%+20% 120% actual selling price. 
100% $60 supposed cost. 
'1. 100%=100% $60, 

2. 1%= T ^ of (100% $60) 1% $.60, and 
!!.<! ^ y> ']3. 25% 25 times (1% $.60) 25% $15 = sup- 
posed gain. [ual selling price. 
(100% $60) + (25% $15) 125% $75 act- 
.; 125% $75 120%, 
5%=$75, 

l%=|of $75 $15, and 
100% 100 times $15 $1500 cost of the goods. 

III. .-. $1500 cost of goods. (Milne's prac., p. 261, prob. 40.) 



82 FINKEL'S SOLUTION BOOK. 

Note, The selling price is the same in the last condition of 
this problem as in the first. Hence we have the selling price in 
the last condition equal to the selling price in the first as shown 
in step (7.) 

I. I sold an article at 20% gain, had it cost me $300 more, I 
would have lost 20%; find the cost. 



II. 



(1- 
(2. 

(3. 



100% actual cost of the article. 

20% actual gain. 
100%+20% 120% actual selling price. 



(4.) 100% +$300 supposed cost. 
1. 100% 100%+$300, 

of (100%+$300) 1%+$3> and 




!!.<! vt "']3, 20% 20 times (1%-f $3) 20% +$60 supposed 
loss. [ual selling price. 

(100% +$300) (20% +$60) 80%+$240=act- 
.-. 120% 80%+$240. 
40% $240, 

1% ^o- of $240 $6, and 
100% 100 times $6 $600 cost of the article. 

III. .-. -$600 cost of the article. 

(7?. H. A., p. 409,prob. 85.) 

I. A man wishing to sell a horse and a cow, asked three 
times as much for the horse as for the cow, but, finding 
no purchaser, he reduced the price of the horse 20%, 
and the price of the cow 10%, and sold them for $165. 
What did he get for each? 

1.) 100% asking price of the cow. 

2.) 300% asking price of the horse. 
3.) 10% reduction on the price of the cow. 

>) 100% 10% 90% selling price of the cow. 



(1. 100% 300%, 
(5.K2. l%=iio of 300% 3%, and 

13. 20% 20 times 3% 60% reduction on horse. 

(6.) 300% 60% 240% selling price ot the horse. 

(7.) 90% +240% 330% selling price of both. 

(8.) $165 selling price of both 

(9.) .-. 330% $165, 
(10.) l%=idny of $165 $.50, and 

(11.) 90% 90 times $.50 $45 selling price of cow. 

(12.) 240% 240 times $.50 $120 selling price of 
horse. 

Ill / $45 amount he received for the cow, and 
' I $120 amount he received for the horse. 






PROFIT AND LOSS. 



EXAMPLES. 

1. What price must a man ask fora horse that cost him $200, 
that he may iali 20% on his asking price and still gain 20% ? 

Ans. $300. 

2. A man paid $150 for a horse which he offered in trade at 
a price he was willing to discount at 40% for cash, as he would 
then gain 20%. What was his^trading price? Ans. $300. 

3. A man gained 20% by selling his house for $3600. What 
did it cost him ? Ans. $3000. 

4. A gained 120% by selling sugar at 8? per pound. What 
did the sugar cost him per pound? Ans. 3 T 7 T /. 

5 How must cloth, costing $3.50 a yard, be marked that a 
merchant mavr deduct 15% from the marked price and still gain 
15%? Ans.W&fr. 

6. Sold a piece of carpeting for $240, and lost 20%; what 
selling price would have given me a gain of 20% ? 

Ans. $360. 

7. Sold two carriages for $240 apiece, and gained 20% on 
one and lo?t 20% on the other; how much did I gain or lose in 
the transaction ? Ans. Lost $20. 

8. Sold goods at a gain of 25% and investing the proceeds, 
sold at a loss of 25% ; what was my % of gain or loss. Ans. 6J%. 

9. A man sold a horse and carriage for $597, gaining by the 
sale, 25% on the horse and 10% on the cost of the carriage. If 
j of the cost of the horse equals f of the the cost of carriage, 
what was the cost of each? Ans. Carriage $270; horse $240. 

10. If ^ of the selling price is gain, what is the profit? 

Ans. 80%. 

11. If -J- of an article be sold for the cost of -J of it, what is 
the rate of loss? Ans. 33^%. 

12.. I sold two houses for the same sum; on one I gained 25% 
and on the other I lost 25%. My whole loss was $240; what 
did each house cost? Ans. First $1440, second $2400. 

13. My tailor informs me that it will take 10i sq. yd. of 
cloth to make me a full suit of clothes. The cloth I am about to 
buy is 1J yards wide and on sponging it will shrink 5% in length 
and width. How many yards will it take for my new suit? 

Ans. 



14. A grocer buys coffee at 15/ per ft>. to the amount of $90 
worth, and sells it at the same price by Troy weight ; find the 
of gain or loss. Ans. Gain 



84 FINKEL'S SOLUTION BOOK. 

15. I spent $260 for apples at $1.30 per bushel ; after retain- 
ing a part for my own use, I sold the rest at a profit of 40%, 
clearing$13on the whole cost. How many bushels did I retain? 

Ans. 50 bu. 

16 How must cloth costing $3.50 per yard, be marked that 
the merchant may deduct 15% from the marked price and still 
make 15% profit? Ans. $4.735. 

17. 1 sold goods at a gain of 20%. If they had cost me $250 
more than they did, I would have lost 20% by the sale- How 
much did the goods cost me? Ans. $500. 

18. A merchant bought cloth at $3.25 per yard, and after 
keeping it 6 months sold it at $3.75 per yard. What was his 
gain %, reckoning 6% per annum for the use of money? 

Ans. 12%+. 



V. STOCKS AND BONDS. 



1. Stocks is a general term applied to bonds, state and 
national, and to certificates of stocks belong to corporations. 

3. A Bond is a written or printed obligation, under seal, 
securing the payment of a certain sum of money at or before a 
specified time. 

3. Stock is the capital of the corporation invested in busi- 
ness; and is divided into Shares, usually of $100 each. 

4. An Assessment is a sum of money required of the 
stockholders in proportion to their amount of stock. 

5. A Dividend is a sum of money to be paid to the stock- 
holders in proportion to their amounts of stock. 

6. The far Value of money, stocks, drafts, etc., is the 
nominal value on their face, 

7. The Market Value is the sum for which they sell. 

8. Discount is the excess of the par value of money, 
stocks, drafts, etc., over their market value. 

9. Premium is the excess of their market value over their 
par value. 

1C. Brokerage is the sum paid an agent for buying stocks, 
bonds, etc. 



STOCKS AND BONDS. 



85 



II. 



I. At |% brokerage, a broker received $10 for making. an in- 
vestment in bank stock ; how many shares did he buy? 

1. 100% par value of stock. 

2. i% brokerage. 

3. $10 brokerage. 

4. .-. i%=$10, 

5. 1% 4 times $10 $40, and 

6. 100% 100 times $40 $4000 par value of stock. 

7. $100 par value of one share. 

8. $4000 par value of 4000-r-lOO, or 40 shares. 

III. . 40 number of shares. 

I. How many shares of railroad stock at 4% premium can 
be bought for $9360? 

rl. 100% par value of stock I can buy. 

2. 4% premium. 

3. 104% price of what I buy. 

4. $9360 price of what I buy. 
II.<!5. .-. 104% $9360. 

6. 1 %=rht of $9360 $90. 

7. 100% 100 times $90 $9000 par value. 
$100 par value of one share. 

^9. $9000 par value of 9000-^-100, or 90 shares. 

III. .-. 90 number of shares that can be bought. 

I. When gold is at 105, what is the value of a gold dollar in 
currency ? 

1. 105X ; or 105% in currency 100/; or 100% in gold. 

2. I/; or 1% in currency .95-^y/ ; or .95-^% in gold. 

3. 100X; or 100% in currency 95^ T X 5 or 95^% in gold. 
.*. $1 in currency is worth 95^X in gold. 

In 1864, the "greenback" dollar was worth only 35f/ in 
gold; what was the price of gold ? 

1. 35fX; or 35f% in gold 100/ ; or 100% in currency. 

2. I/; or 1% in gold -^ of 100X; or 100% 2.8X ; or 
2.8% in currency. }O T 

3. 100X; or 100% in gold 100 times 2.8X; or 2.8% 280/; 

or 280% in currency. 

/. $1 in gold was worth $2.80 in currency. 

(R. 3d p., p. 217,prob. 8.) 



II. 

Ill, 
I. 



Bought stock at 10% discount, which rose to 5% premium 
and sold for cash. Paying a debt of $33, I invested the 
balance in stock at 2% premium, which at par, left me 
$11 less than at first; how much money had I at first? 



86 FINKEL'S SOLUTION BOOK. 

(1.) 100%=my money at first. 

(2.) 100%=par value of stock. 

(3.) 10%=discount. 

(4.) 100% 10%=90%=market value. 

(5.) .'. 90%=100%, my money; because that is the 

amount invested. 

(6.) l%=rv of 100%=1%, and 

(7.) 100%=100 times l$,=lll%=par value of the 

stock in terms of my money. 



(8.) 



II. 



/q \ 



2. 1% l-|f%,and [terms ot my money. 

3. 5% 5 times 11% 5-f %=premium on stock in 

4. llll%-|-5-t%=116f% = what I received for the 

stock. 

5. 116f % $33=amount invested in second stock. 

1. 100%=par value of second stock. 

2. 2%=premium. 

3. 100%-f-2%=102%=:market valueof second stock. 

4. 116f% $33=market value of second stock 

5. .-. 102%=116J % $33, 

6. 1%=^ of 

7. 100%=100 times 

$32 T 6 T =par value of second stock. 

(10.) 114 T \\% $32 T 6 T =what 1 received for the second 

stock, since I sold them at par. 
(11.) .'. 114 T Vij% $32 T 6 T =100% $11, by the last con- 

dition of the problem. 
(12.) 1 



(13.) i%__L^ of $21 T 6 T =$1.485, and 
14 T %% 

(14.) 100%=100 times $1.485=$148.50. 
III. .-. I had $148.50 at first. (7?. H. A., p. 212, prob. 8.) 

I. Bought $8000 in gold at 110%, brokerage % ; what did 

I pay for the gold in currency? 
'1. 100%=par value of gold. 

2. 110%=market value. 

3. -%=brokerage. 

H.<[4. 110%+4%=110i%= entire cost. 

5. 100% $8000, 

6. 1%=!-^ of $8000=$80, and 

7. 110l%=110^ times $80=$8810=cost of gold in currency. 

III. .'. $8000 in gold costs $8810 in currency. 

I. What income in currency would a man receive by invest- 
ing $5220 in U. S. 5-20, 6% bonds at 116%, when gold 
is worth 105? 



STOCKS AND BONDS. 87 

100%=par value of the bonds. 

116% market value. 
$5220 market value. 
.-. H6% =$5220. 

1% T | of $5220 $45. 

100% 100 times $45 $4500=par value of bonds. 
100% $4500. 

1%^ of $4500 $45. 

6% 6 times $45 $270 income in gold. 
$1.00 in gold $1.05 in currency. 
$270 in gold 270 times $1.05 $283.50 in currency. 

$283.50 income in currency. 

(/?. 3d p., p. 217,prob. 5.) 



What % of income do U. S. 4^-% bonds, at 108, yield when 
gold is 105%? 

(1.) 100% amount invested in the bonds. 

(2.) 100% par value of bonds. 

(3.) 108%=rnarket value. 

(4.) .-. 108% 100%, from (1). 

(5.) l%=riir of 100% ff%, [of amount invested. 

TT I (6.) 100% 100 times |-f-%=92$% par value in terms 
100% 92-J%. 

4% 4^ times ff%=4-^% income in gold. 
8.) 100% in gold 105% in currency. 
9.) 1% in gold T -^ of 105% l-fa% in currency. 

^(10.) 4-g-% in gold ig- times l- 2 - 1 ^% 4f % in currency. 

III. .-. Income in currency 4f %. 

Note. This is a general solution of the preceding problem. 
Since there is no special amount given, we represent the amount 
invested by 100%. The market value and the amount invested 
being the same, we have 108% 100% as shown in (4). 

I. A man bought Michigan Central at 120, and sold at 124%;. 
what % of the investment did he gain? 




II.< 



1. 124% selling price. 

2. 120 cost. 



3. 124% 120% 4% gain. 

4. 120% 100% of itself. 

5. l%=ThfOf 100%=f%, 

6. 4% 4 times |%=3%=gain on the investment. 



III. .. He gained 3% on the investment. 



I. 



II. 



FINKEL'S SOLUTION BOOK. 

What sum invested in U. S. 5's of 1881, at 118, yielded an 
annual income of $1921 in currency, when gold was 
at 113? 




$1.13 in currency=$l in gold. 

$1 in currency^^Vg. of $l=$i^| in gold, and 

$1921 in currency=1921 times 

come in gold. 
100%=par value of the bonds. 

5 % income in gold. 
$1700=income in gold. 
.-. 6%= $1700, 

l%==\ of $1700=$340, and [bonds. 

100%=iOO times $340=$34000=par value of the 
100%=$34000, 

l%=^f of $34000=1340, and 
118%=118 times $340=$40120=market value, or 

amount invested. 



III. .'. $40120=amount invested. 



II. 



III. 



I. 



(I-) 
(2. 

(3. 

(4- 
(5. 
(6. 



(7.) 



(8.) 



(9.) 
(10.) 



(11.) 
(12.) 



SECOND SOLUTION. 

100%=amount invested in currency. 
100%=par value. 
118%=market value. 
... 118%=100%, from (1.) 

l%=rh- of 100?fc=HHF%> and 

100%=100 times f|% = ^84||%=par value in 
terms of the investment. 



and 

d%=5 times f %==4^%==income in gold. 
100% in gold 113% in currency, 



in gold=l- r '^ n currency 
in gold=4i| times 
n currency. 
$1921=income in currency. 



and 




of $1921=$401.20, and 

100%=100 times $401.20=-$40120^amount in- 
vested in currency. 

$40120=amount invested. (JR. 3d p., p. 218, prob. 8.) 



How many shares of stock bought at 95^%, and sold at 
105, brokerage \% on each transaction, will yield an 
income of $925 ? 



STOCKS AND BONDS. 



II. 



1. 100%=par value of stock. 

2. 95J%=market value of stock. 

3. ^%==:brokerage. 

4. 95i+i%=95i%=entire cost. 

5. 105%=selling price-{-brokerage. 

6. ^%=brokerage. 

7. 105% i%= 1041%=selling price. 

8. 104|% < 

9. $925=gain. 

10. .-. 9i% : 

11. 1 %=~ of $925=$100, and 

ft* 

12. 100%=100 times $100=$10000==pai value of stock. 

13. $100=par value one share. 

14. $10000=par value 10000-f-lOO, or 100 shares. 



III. .-. 100=number of shares. 



(R. 3d p., p. 218,prob. 9.) 



II. 



If .1 invest all my money in 5% furnace stock salable at 
75%, my income will be $180; how much must I bor- 
row to make an investment in 5% state stock selling at 
102%, to have that income? 

1. 100%=par value of furnace stock. 

2. 5%=income. 

3. $180=income. 

4. .-. 5%=$180, 

5. l%=i of $180=$36, and [nace stock. 

6. 100% =100 times $36=$3600=par value of fur- 

1. 100%=$3600, 

2. 1%= T ^ O t $3600=$36, and [nace stock. 
$. 75%=75 times $36=$2700=market value of fur- 

1. 100%=par value of state stock. 

2. 6%=income. 



(!) 



(8.) 



(5.) 



3. $180=income. 

4. .-. 6%=$180, 

5. 1%=^ of $180=$30, and [stock. 

6. 100%=100 times $30=$3000=par value of state 

1. 100% =$3000, 

2. 1%= T ^ of $3000=$30, and [state stock. 

3. 102%=102 times $30=$3060=market value of 
$3060 $2700=$360=what I must borrow. 



III. .-. I must borrow $360. 



(R. H. A., p. 225, prob. 2. 



When U. S. 4% bonds are quoted at 106, what yearly in- 
come will be received in gold from bonds that can be 
bought for $4982 ? 



90 FINKEL'S SOLUTION BOOK. 

(1.) 100% par value of the bonds. 
(2.) 106% market value. 

(3.) $4982 market value, or amount invested. 
I I (4.) .-. 106% $4982, 

(5.) 1 % yfo of $4982 $47, and 

(6.) 100% 100 times $47 $4700. 

r l. 100% $4700, 
(7.K2- 1% T i<5- of $4700 $47, and 

13. 4% 4 times $47 $188=income in gold. 
JII. .-. $188 income in gold. (fi. 3p., p. 218, prob. 11.) 

I. The sale of my farm cost me $500, but I gave the pro- 
ceeds to a broker, allowing him ^%, to purchase rail- 
road stock then in the market at 102%; the farm paid 
5% income, equal to $2075, but the stock will pay 
$2025 more; what is the rate of dividend? 

(1.) 100% value of the farm. 

(2.) 5% income on the farm. 

(3.) $2075 income on the farm. 

(4.) .-. 5% $2075, 

(5.) l%=i of $2075 $415, and 

(6. ) 100% 100 times $415 $41500 value of farm. 

(7.) $41500 $500 $41000 amount invested in stock. 
1. 100% par value of the stock. 

2. 102% market value, or amount invested. 

3. ^% brokerage 

4. 102%+^% 102|% entire cost of stock. 

fc=$41QOQ, 



of $41000 $400, and 

[railroad stock. 
7. 100% 100 times $400 $40000 par value of the 

1. $2075+$2025 $4100 income on railroad stock. 

2. $40000100%, 

3. $1%=^^ of 100%^% , and [dend. 

4. $41004100 times =W= rate of divi- 



III. /. 10^% rate of dividend. (/?. H. A., p. 224., prob. 4-) 

I. What must be paid for 6% bonds to realize an income of 
8% on the investment? 

1. 100% amount invested. 

2. 6% income on the par value of the bonds. 

3. 8 % income on the investment. 

4. .'. 8% of investment 6% of the par value, 

5. 1% of investment^ of 6% f % of the par value, and 
100% of investment 100 times f% 75% of par value. 

III. .-. Must pay 75% to make 8% on the investment. 

Note. It must be borne in mind that 100% of any quantity is 
the quantity itself. .-. 100% of the amount invested equals the 



II. 




STOCKS AND BONDS, 



91 



II. 






II. 



amount invested. It must also be remembered that the income 
on the par value is equal to the income on the investment. Sup- 
pose I buy a 500-dollar 6% bond for $400. The income on the 
par value, or face of the bond is 6% of $500, or $30. But $30 is 
7%% of $400, the amount invested. Hence, the truth of step 4 
in the above solution. 

Which is the better investment, buying 9% stock at 
25% advance, or 6% stock at 25% discount. 

1.) 100%amount invested in the 9% stock, 
100%=par value. 
25%==premium. 

100%+25%=125%= market value. 
.-. 125%=100%, 

1%=^ of 100%=f %, and 
100%=100 times f %=80%=par value in terms 
of the investment. 

1. 100%=80%, 

2. I%=TO-H of 80%=|%, and [stock. 

3. 9%=9 times |%=:7i#==income of 9% 
100%=amount invested in 6% stock. 
100%= par value of 6% stock. 

25%=discount. 

100% 25%=75%=market value. 
.-. 75%=100%. 

1%= T V of 100%=li%, and 

100 times l^%=133^%=par value of 
the 6% stock in terms of the investment. 



L B. 



2.) 
(3. 
(4- 
(5. 
(6.) 
(7-) 



(8.) 



(2.) 
(3.) 
(4-) 
(5.) 
(6.) 
(7.) 



(8.)J2. 1%= 
IS. 6%= 



and [stock. 

6%=6 times l%=8%=income of 6% 
III. .-. The latter is the better investment, since it pays 8% 
7.1.% ? or \ c jc more income on the investment. 

( Greenleafs N. A., p. 298, prob. 5.) 

If I pay 87i% for railroad bonds that yield an annual in- 
come of 7%, what % do I get on my investment? 
100%=investment. 
100%=par value. 

87^%=market value, or amount invested. 
.-. 87%=100%, from (1.) 

1% 1- of 100%=1|%, and 

o/'ff 

100%=100 times l^%=114^%==par value in 
terms of the investment. 

1. 100%=114f%, 

2. 1%== T ^ 15 . of 114f %=1^%, and [ment. 

3. 7%=7 times l|%=8%=income on the invest- 



(I-) 

(2.) 

It! 

(5.) 
(6.) 



(7.) 



Ill. 



)=income on the investment. 



92 



FINKEL'S SOLUTION BOOK. 



A banker owns 2-J% stocks at 10% below par, and 3% 
stocks at 15% below par. The income from the former 
is 66f % more than from the latter, and the investment 
in the latter is $11400 less than in the former; required 
the whole investment and income. 



II. 



!J 


1. 




'2. 


(3.). 


4. 




5. 




6. 




rl. 


( 4 ,)I: 




L 
1. 




2. 




3. 


(5.)< 


4. 
5. 




6. 




1. 


/A J* 


( 6 -) 3. 



(8.) 

(9.) 

(10.) 

(11.) 

(12.) 

(13.) 

(14.) 



100%=investment in the former. 

100% $11400=investment in the latter. 

100%=par value of the former. 
10%=discount of the former, [vested in former., 

100% 10%=90%=market value, or amount in- 

.-. 90%=100%, from (1), 
1%=-^ of 100%=!-^%, and 

100%=100 times l%=lll%=par value of for- 
mer in terms of the investment. 

100%=11H%, 

2-J-%=2 times 11% 2J%=income of former in 

terms of the investment. 
100%=par value of the latter. 

15%=discount. [vested in the latter. 

100% 15%=85%=market value, or amount in- 
.-. 85%=100% $11400, from (2), 

1%=^ of (100% $11400)=1 T 3 T % $134 T 2 T , 
100%=100 times (1 T 3 T % $134 T 2 7 ) = 117jj% 

$13411^- 3 r=par value of latter in terms of former. 

100%=117||% $13411if, [$134 T 2 T , and 

1% = T ^ of (H7fi.% $l3411f 3 - )= l T 3 y % 
3%=3 times (lfV% $134 1 2 T )=3 1 9 T % $402 T 6 T 
=income of latter in terms of the investment. 
100%=income of the latter. 
100%+66|%=166|%=income of the former. 
2J%=income of the former. 
/. 166f %=2^%, 

of2|%=Jo%^nd 

[terms of income of former. 



"166* 



100%=100 times ^%=lj%=income of latter in 
3 T 9 T % $402 T 6 T ==income oif the latter. 



, 
Iff [former. 

100%=100 times $216=$21600=investment in 
100% $11400= $21600 $11400 =$10200=- in- 
vestment in latter. 

=2^- times $216=$600=income of former. 



STOCKS AND BONDS. 



III. 



I. 



II. 



(15.) aA-% $402;fr=3JV times $216 $402- 1 <y=$36Q= 

income of latter. 

(16.) $21600+$10200=$31800=whole investment 
(17.) $600+$360=$960= whole income. 

J$31800=whole investment, and 
' |$960=whole income. (/?. H. A., p. 225, prob. 4. ) 

W. F. Baird, through his broker, invested a certain sum 
of money in Philadelphia 6's at 115^-%, and three times 
as much in Union Pacific 7's at 89-J%, brokerage -j-% in 
both cases; how much was invested in each kincf of stock 
if his annual income is $9920? 

(1.) 100%=amount invested in Philadelphia 6's. 
(2.) 300%=amount invested in Union Pacific 7's. 





i. 
2. 


iuuy0=pai vaiue 01 .rmiauejpma. o s>. 
115%=market value. 




3. 


-j-% brokerage. 


(3-). 


4. 

5. 


115-J%+|%=ll"6%=entire cost of Phila. 6's. 
.r. 116%=100%. 




6. 


1%=^ of 100%=||%, and 




7. 


100% 100 timesff %= 86A%=par value of Phil- 






adelphia 6's in terms of investment. 


rl. 


100% =86 2^%, 


MS 


1 % .j-i-g. of 86^-% , and 
6%=6 times ff %=5 T 5 %=income of Philadel- 


I 


phia 6's in terms of investment. 




1. 


100%=par value of Union Pacific 7's. 




2. 


89^%=market value. 




3. 


^.% brokerage. 


/ r \ 


4. 


89|%-}-^==90%==entire cost of Union Pacific 7's. 


(&r 


5. 


... 90%=300%, 




6. 


l%=fa of 300%=3%, and 




7. 


100%=100 times 3|%=333i%=par value of 






Union Pacific 7's. 


rl. 


100%=333%, 


(6.)j 2 - 


-j (ji i Q.C QQQ i <T/. Q JL cL. and 


1 


7%=7 times 3J%=23^%=income of Union Pa- 



(7.) 
(8.) 
(9.) 

(10.) 

(11.) 
(12.) 



cific 7's in terms of investment. 

52^+ 23 i% =28 !r%= whole income. 
$9920=whole income. 



- f ^ 9920 



[in Philadelphia 7's. 
100%=100 times $348=$34800=amount invested 
300%=300 times $348=1104600=5= amount in- 
vested in Union Pacific 7's. 



FINKEL'S SOLUTION BOOK. 



III. 



I. 



II. 



( $34800=a mount invested in Philadelphia 6's, and 
I $104400 amount invested in Union Pacific 7's. 

(R. H. A., p. 225,prob. 6.) 

Thomas Reed bought 6% mining stock at 114^%, and 
4% furnace stock at 112% , brokerage -% ; the latter 
cost him $430 more than the former, but yielded the 
same income ; what did each cost him? 



(2.) 



(3.) 



(4.) 



100%=amount invested in mining stock. 
100%+$430=amount invested in furnace srock. 

1. 100%=par value of mining stock. 

2. 114^%=market value. 

4. ii4^%^-^%=1ilo%=ent\re cost. 

5. /. 115^=100%, from (1), 

6. l%= T l-g of 100% ff %, and 

7. 100%=100 limes |J%=965|^=par vnlue oi 

mining stock in terms of investment. 



(5.) 



(6.) 

(7.) 
(8.) 



l%=rb of 96ff %=$%, and 
6%=6 times |^%:==5-2\% 

stock in terms of investment. 
100%=par value of furnace stock. 
112%=market value. 



of 



mining 



=~. T of (100%+$430)=f%+$3it, and 
-- 



100%=100 times (|%-|_$3|l)_88|%-|-$382f= 
par value of furnace stock in terms uf inv r estm't. 



2. l%= r J T of (88f%+$382| )=|%+|3|l, and 

3. 4%=4 times (|%+$3H)= 3|- %+$15i|=income 

of furnace stock in terms of the investment. 
.-. 5A%= =s 3f%.-fll5tti b y the conditions of the 
problem, 



!a of $15= 



and 



(10.) 
(11.) 



[mining stock. 
100%=100 times $9,20=$920=amount invested in 
100%+|430=^1350=amount invested in furnace 
stock. (^?. H. A., p. 225,prob. 7.) 



Ill * /$920 amoun t; invested in mining stock, and 
' *|$1350=amount invested in furnace stock. 



STOCKS AND BONDS. 



95 



I. 



n. 



(i.) 



Suppose 10% state stock is 20% better in market than 4% 
railroad stock; if A.'s income be $500 from each, how 
much money has he paid for each, the whole investment 
bringing 6-gf 3-% ? 

1. 100 %=par value of state stock. 

2. 10%=income. 

3. $500= in come. 

4. .-. 10% =$500, 

5. 1%= T L of $500=$50,and [stock. 

6. 100%=100 times $50=$5000=par value of state 

1. 100%=par value of railroad stock. 

2. 4%=income. 

3. $500=income. 

4. .-. 4%=$500, 

5. l%=i of $500=$125, and [railroad stock. 

6. 100%=100 times $125=$12500=par value of 
$5000=f of $12500, i. e., the face of state stock 

is |- of face of railroad stock. 

1. 100%=whole investment. 

2. G^l-g %=in come of whole investment. 

3. $500+$500=$1000=income of whole investment. 



(2.) 



(3.) 



(4.) 



(5.) 



6. 



1%= L- of $1000=$166.50, and 

6^ [ment. 

100%=100 times $166.50=$16650=whole invest- 



1. 100%=investment in railroad stock. 

l'. 40%=| of 100%=investment in state stock, 

excluding the 20% excess. 
2'. 100% =40%, 
3'. l%= T ^o of 40%=|%, and 
4. 20%=20 times |% =8%=excess of state 

stock over same amount of railroad stock. 

3. 40%+8%=48%=investment in state stock. 

4. 100%+48%=148%=whole investment. 

5. $16650=whole investment. 

6. .-. 148%=$16650, 

7. l%= T ^g- of $16650=$112.50, [railroad stock. 

8. 100%=100 times $112. 50=$11250=investment in 

9. 48%=48 times $112.50=$5400=investment in 

state stock 

j. ^ J$11250=amount invested in railroad stock, and 
' ' ' 1 $5400=amount invested in state stock. 

(R. H, A,, p. 227, prob. 5.) 



9d FINKEL'S SOLUTION BOOK. 



EXAMPLES. 

1. What could I afford to pay for bonds yielding an annual 
income of 9% to invest my money so as to realize 6% on the in- 
vestment? Ans. 150%. 

2. What must I pay for Chicago, Burlington & Quincy Rail- 
road stock that bears 6% that my annual income on the invest- 
ment may yield 5% ? Ans. 120%. 

3. Bought 75 shares N. Y., P. & O. Railroad stock at 105%, 
and sold them at 108% ; how much did I gain in the transac- 
tion? Ans. $262.50. 

4. How many shares of bank stock at 5% premium, can be 
bought for $7665? Ans. 73. 

5. A broker bought stock at 4% discount, and, selling them 
at 3% premium, gained $1400 ; how many shares did he buy? 

Ans. 200. 

6. At what price must I buy 15% stock that it may yield the 
same income as 4% stock purchased at 90% ? Ans. 337-^%. 

7. How much must I pay for New York 6's so that I may 
realize an income of 9%? Ans. 66f %. 

8. At what price must I buy 7% stock so that they may yield 
an income equivalent to 10% stocks at par? Ans. 70%. 

9. What sum must I invest in U. S. 6's at 118% to secure an 
annual income of $1800 ? r Ans. $35400. 

10. Which is the more profitable, and how much, to invest 
$5000 in 6% stock purchased at 75%, or 5% stock purchased at 
60%? Ans, The latter; $16|. 

11. If a man who had $5000 U. S. 6's of 1881 should sell them 
at 115%, and invest in U. S. 10-40' s purchased at 105%, would 
he gain or lose and how much? Ans. Loss $26.19. 

12. When gold is at 120, what is a "greenback" dollar worth ? 

Ans. 



13. Suppose the market value of 5% bank stock to be 'll-g-% 
higher than 8% corporation bonds ; I realize 8% on my invest- 
ment, and my income from each is $180; what did I invest in 
each? Ans. $2923.07 A in former, and $1576.92 T \ in latter. 

14. A bought 5% railroad stock at 109|%, and 4|% pike stock 
at 107-J%, brokerage |% ; the former cost $100 less than the latter 
but yielded the same income; what did each cost him? 

Ans. $1100 cost of former, and $1200 cos of latter 



INSURANCE. 97 

15. What rate % of income shall I receive if I buy U. S. 5's at 
a premium of 10%, and receive payment at par in 15 years? 

Ans. " 



16. Suppose the market value of 6% corporation stock is 20% 
less than 5% state stock; if my income be $1200 from each, what 
did I pay for each if the whole investment brings 6% ? 

Ans. $16000, and $24000. 



17. I bought 2|% stock at 80%, and 4|% stock at 
The income on the former was 44f % more than on the latter, 
but my investment is $22140 less in the latter than in the former; 
what do I realize on my investment? Ans. 






Hint. Find the whole investment, and whole income as in the problem 
on page 75. Then find what % the whole income is of the whole invest- 
ment. 

18. Invested in U. S. 4's at 105, brokerage % ; f as much 
in U. P. 6's at 119|, brokerage %; and 3 times as much in N. 
Y. 7's, at 87J, brokerage \%. If my entire income is $1702, find 
my investment. (School Visitor, vol. 12, p. 97.) Ans. $25320. 

19. A. paid $1075 for U. S. 5-20 6% bonds at 7|% premium, 
interest payable semi-annually in gold. When the average pre- 
mium on gold was 112%, did he make more or less than B. who 
invested an equal sum in railroad stock at 14% below par, which 
paid a semi-annual dividend of 4%? 

Ans. A. makes $16.40 less than B. every six months. 

20. I invested $4200 in railroad stock at 105, and sold it at 
80%; how much must I borrow at 4% so that by investing all I 
have in 6% bonds at 8% interest, payable annually, I may re- 
trieve my loss in one year? Ans. $18600. 



VII. INSURANCE. 

1. Insurance is indemnity against loss or damage. 

~ T M. Fire Insurance. 

1. Property Insurance, j 2 j.^ Insurance 



2. Insurance. 



2. Personal Insurance. 



1. Life Insurance. 

2. Accident Insurance. 



3. Health Insurance. 

3. Property Insurance is the indemnity against loss or 
damage of property. 

4. Personal Insurance is indemnity against loss of life 
or health. 

5. Fire Insurance is indemnity against loss by fire. 



98 FINKEL'S SOLUTION BOOK. 

6. Marine Insurance is indemnity against the dangers 
of navigation 

7. Life Insurance is a contract in which a company 
agrees, in consideration of certain premiums received, to pay a 
certain sum to the heirs or assigns of the insured at his death, or 
to himself if he attains a certain age. 

8. Accident Insurance is indemnity against loss by 
accident. 

9. Health Insurance is a weekly indemnity in case of 
sickness. 

1C. TJie Insurer, or Underwriter, is the party, or 
company, that undertakes the risk. 



11. The Itisfc is the particular danger against which the 
insurer undertakes. 

12. The Insured is the party protected against loss. 



13. The frennium is the sum paid for insurance; and is 
a certain per cent, of the amount insured. 

14. The Amount, or Valuation, is the sum for which 
the premium is paid. 



I. My house is permanently insured for $1800, by a deposit 
of ten annual premiums, the rate per year being |%; 
how much did I deposit, and if, on terminating the in- 
surance, I receive my deposit less 5% ; how much do I 
get? 

(1.) 100%=$1800, 

(2.) I%=TT><T of $1800=$18, and 

II. 



\&. ) j. /o :== ^-^ wi tpouu=<j>io, anu 

(3.) i%=| times $18=$13.50=one annual deposit. 

(4.) $135=10 times $13.50=ten annual deposits. 

/I. 100%=$135, 
(5.){2. 1%=^ of $135=$1.35, and 

13. 5%=5 times $1.35=$6.75=deduction. 
(6.) $135 $6.75=$128.25=what I received. 

ry . ( $135=amount deposited, and 
' ' ' ( $128.25=amount received. 

(7?. H. A., p. 230, prod. 5.} 

I. An insurance company having a risk of $25000, at T 9 ^%, 
reinsured $10000, at \% , with another office, and $5000, 
at 1%, with another; how much did it clear above what 
it paid ? 



INSURANCE. 



99 



II. 



(1.) 100% $25000, 

(2.) l%=T-b of $25000 $250, and 

(3.) T 9 (r%=T 9 <r times $250 $225 what the company 
received for taking the risk. 

1. $10000 amount the company reinsured at f%. 

2. 100% $10000, 

(4.) 3. 1%= -rfoof $10000 $100, and 

4. f% f times $100 $80 what the company paid 
for reinsuring $10000. 

1. $5000 amount reinsured in another office at 1%. 

2. 100% $5000, [for reinsuring $5000. 
(5.) 3. 1% T |-g- of $5000 $50 what the company paid 

4. $80+$50 $130 what the company paid out. 

5. $225 $130 $95 what it cleared. 

III. .-. $95 what the company cleared. 

(/?. H. A., p. 230,prob. 7.) 

I took a risk at 4-J% ; reinsured -| of it at 2%, and of it 
at 2-^% ; what rate of insurance do I get on what is left? 



(1.) 

(2.) 



100% whole risk. 

premium. 

| of 100%^amount reinsured at 2%. 



? y^-g- of 40% 1%, and [suringf of therisk. 
% 2 times -| % 1^% ampunt I pay out for rein- 
fl. 25% i of 100% second part reinsured. 
2. 100% 25%. 
(4.K8. 1% T ^r of 25% i%, and 

4. 2^-% 2-J- times ^%|% amount I paid out for 
( reinsuring of the risk. 

(5.) |%-|-f%l^J% amount of premiums paid out. 
(6.) 11%1^%^%amountof premium I had left. 
(7.) 40%-f-25% 65% whole amount reinsured. 
(8.) 100% 65% 35% risk left on which I received 

TQ% premium. 
f l. 35% 100% of itself. 
(9.K2, 1% ^ of 100% 2f%, and 

(.3. ^ofa.^ times 2^%= T 3 - ? .%=r=rate of premium 

III. ,'. T 3 ? % rate of insurance I receive. 

(R. H. A., p. 281,prob. 6.) 

Remark. 35% is the base and -fa% is the percentage, and we 
wish to know what per cent, -}$% is of 35%. 

I. Took a risk at 2%; reinsured $10000 of it at 2^% and 
$8000 at 1|%; my share of the premium was $207-50; 
what sum was insured? 



100 FINKEL'S SOLUTION BOOK. 

rl. 100%=$10000, 

(1.K2. 1%= -r^of $10000=4100, and [$10000 reinsured. 
13. 



r%=2^ times $100=$212.50=amount paid out on 

rl. 100%=$8000, 

(2.K2. l%= T i of$8000=$80,and [$8000 reinsured. 

13. If %=lf times $80=$140=amount paid out on 

(3.) $212.50+$140=$352.50=whole amount paid out. 

(4.) $207. 50=what I realize. 

(5.) .'. $352.50+207.50=$560=premium on whole risk. 

(6.) 100%=risk. 

(7.) 2%=premium. 

(8.) $560=premium. 

(9.) .'. 2%=$560, 

(10.) l%=i of $560=$280, and 

(11.) 100^=100 times $280=$28000=risk. 

III. .-. $28000=risk. (R. H. A., p. 232, prob. 6.) 



I. I can insure my house for $2500 at T 8 % premium annually, 
or permanently by paying down 12 annual premiums; 
which should I prefer, and how much will I gain by it 
if money is worth 6% per annum to me? 

r (1.) 100%=$2500. 
(2.) l%=Tta of $2500=$25, and 
(3.) -nr%=T 8 times $25=$20=one annual premium. 
(4.) $240=12 times $20=twelve annual premiums. 

1. 100%=the amount that will produce $20 an- 

nually at 6%. 

2. 6%=iriterest. 
!!.<! ft . J3. $20=interest. 

l '^4. /. 6%=$20, 

5. l%=iof$20=$3i, and 

6. 100%=100 times $3i=$333^=the amount that 

will produce $20 annually at 6%. 

(6.) $333 1 H-$20=$353 1 i=amount j wou ld have to pay 
down by the former condition. [tion. 

L (7.) .', $353i~$240=$113i=gain by the latter condi- 

TTT ( The latter is the better. 
' I $113i=gain. 

Remark. In (6) we add $20, since a payment must be made 
immediately. $333 will not produce that sum until the end of 
the year. 



INSURANCE. 



The Mutual Fire Insurance Company insured a building 
and its stock for f of its value, charging If %. The 
Union Insurance Company relieved them of of the 
risk, at l-J-%. The building and stock being destroyed 
by fire, the Union lost $49000 less than the Mutual; 
what amount of money did the owners of the building 
and stock lose? 

100%=value of the building and stock. 
66f%=f of 100%=amount insured. 
lj%=rate of insurance. 
100%=66f%, 

nnr of 66f %=f 
=1| times f%= 

from the owners of the building and stock. 
16f %=i of 66|%=amount of which the Union 
relieved the Mutual. 



= and 



r (i-) 

SI 




and 
=what Mutual received 




(10.) 

(11.) 

(12.) 
(13.) 

(14.) 
(15.) 

(16.) 

(17.) 
(18.) 




(21.) 



of 
1-J%=1-J. times -|-% ^% what the Mutual paid 

the Union for taking the risk of 16|%. 
16f %+li%=17-f%=whole amount the Mutual 

received. [paid out. 

amount the Mutual 
i ^,~= amount the Mutual 

lost." ' 

16f %=amount the Union paid the Mutual. 
^%=amount the Union received from the Mutual. 
.'. 16f % i%=16 T 5 T %=amount the Union lost. 
49 1 V% 16 T \%= 32f%=what the Mutual lost 

more than the Union. [Union. 

$49000=what the Mutual lost more than the 
... 32f %=$49000, 

1 %= JL of $49000=$1500, and 
' 32f [ing and stock. 

100%=100 times $1500=$ 150000= value of build- 

66|%=66|- times $1 500=$ 100000=a mount in- 
sured, [ers lost, it not being insured. 

33j-%=33 times $1500=$50000=what the own- 

100% =$100000, 

1%=-^ of $100000=$1000, and 

l|%=l| times $1000=$1750=what the owners 
paid the Mutual for insurance. 

.-. $50000 + $1750= $51750=whole amount the 
owners lost. 



III. .'. The owners of the building and stock lost $51750. 



102 FINKEL'S SOLUTION BOOK. 



EXAMPLES. 

1. At l-f%, the premium for insuring my stare was $89.10; 
what was the amount of the insurance? Ans. $6480. 

2. The premium for insuring a tannery for f of its value, at 
, was $145.60; what was the value of the tannery? 

Ans. $11648. 

3. A store and its goods are worth $6370. What sum must 
be insured, at 2%, to cover both property and premium? 

Ans. - 

4. The premium for insuring $9870 was $690.90 ; what was 
the rate? Ans. 7%. 

5. A merchant whose stock of goods was valued at $30000, 
insured it for f ot its value, at f %. In a fire he saved $5000 of 
the goods. What was his loss? What was the loss of the in- 
surance companies? Ans. 

6. A man paid $180 for insuring his saw mill for f- of its 
value at 3%; what was the value of the mill? Ans. . 

7. A house which has been insured for $3500 for 10 years, at 
\/c & year, was destroyed by fire ; how much did the money re- 
ceived from the company exceed the cost of premiums? 

Ans. . 

8. Took a risk on a house worth $40000, at 2%; reinsured 
J of it for 2^%, and ^ of it at 2-J%; in each case the amount cov- 
ers premium; how much do I gain? Ans. $99.558* 

9. Took a risk at If %; reinsured f of it at 2J% ; my share 
of the premium was $43 ; what was the amount of the risk? 

Ans. $17200. 



10. Took a risk at 2% ; reinsured -J of ?.t at a rate equal to 
3% of the whole, by which I lost $37.50. What was the value 
of the risk? Ans. $5000. 




SIMPLE INTEREST. 103 

CHAPTER XIII. 

INTEREST. 

I. SIMPLE INTEREST. 

1. Interest is money paid by the borrower to the lender 
for the use of money. 

2. The Principal is the sum of money for which interest 
is paid. 

3. The Rate of interest is the rate per cent, on $1 for a 
certain time. 

4. The Time is the period during which the money is on 
interest. 

5. The Amount is the sum of the principal and interest. 

6. Simple Interest is interest on the principal only. 

7. Legal Interest is at the rate fixed by law. 

Usury is interest at a rate greater than that allowed by- 



8. 
law. 



Let P=the principal, 

r=the interest on $1 for one year, 
7?=l-[-r=a mount of $1 for one year, 
7Z=the number of years, 
yl=amount of P for n years, 
/V=simple interest on P for a year, 
Pnr simple interest on P for n years. 
P -\-Pnr P (l-|-?zr)=amountof P forn years, 
of P for n years. 

(I.); 



~\+nr* * v ' 
... Pnr=AP. 

A P Interest 




When any three of the quantities A, P, n, r are given, the 
fourth may be found. 

CASE I. 
rPrincipal,^ 

Given< Rate, and >to find the interest. Formula, 7=s/V. 
iTime, J 



104 FINKEL'S SOLUTION BOOK'. 

I. Find the interest of $300 for two years at 6%. 
By formula, 

Interest /V=$300X-06x2=$36. 
By 100% method. 

rl. 100%=$300, 

1 2. l%= T fc> of $300=$3, and. 

13. 6%=6 times $3=$18=interest for one year. 
U. $36=2 times $18=interest for 2 years. 

III. .'. $36=interest on $300 at 6% for 2 years. 

CASE II. 

(Principal, >> A/> 

Given<Rate, and >to find the time. Formula, n= = . 

llnterest, ) Pr 

I. In what time, at 5%, will $60 amount to $72? 
By formula, 

AP $72 $60 



By 100% method. 

1. $72=amount. 

2. $60=principal. 

3. $72 $60=$12=interest for a certain time. 
II.<J4. 100% =$60, 

5. i% == _|^ of $60=$f, and 

6. 5%=5 times $f=$3=interest for one year. 

7. $12=interest for 12-4-3, or 4 years. 

III. V. $60 at 5% will amount to $72 in 4 years. 

CASE III. 

rPrincipal,^ AP 

Given^ Time, and >to find the rate. Formula, r= - . 

llnterest, ) 
I. I borrowed $600 for two years and paid $48 interest; what 

rate did I pay? 
By formula, 

AP I 



By 100% method. 
'1. $48=interest for 2 years. 
2. $24= of $48=interest for 1 year. 
II.J3. $600=100%, 

4. $l=:^oflOO%=i 
[5. $24=24 times i-%=4 
III. .'. I paid 4% interest. 



SIMPLE INTEREST: 105. 



rTime, ^ 

Given<[Rate, and >to find the principal. 
llnterest. J Fo 



CASE IV. 
rTime, 

A P 1 
Formula, P== 

I. The interest for 3 years, at 9%, is $21.60; what is the 
principal ? 

By formula, 

P _A-P_ 7_$21.60 . 
~-~-- 



II. 



By 100% method. 

1. $21.60=interest for 3 years. 

2. $7.20=4 of $21.60=interest for 1 year. 

3. 100%=principal. 

4. 9%=interest for 1 year. 



5. $7.20=interest for 1 year. 

6. .-. 9%=$7.20, 

7. l%=i of $7.20=$.80, and 
100%=100 times $.80==$80=principal. 

III. .'. $80=the principal. 



CASE V. 
Given<(Rate, and>to find the principal. Formula, P= 



f Time, i A 

LAmount J l + nr ' 



I. What principal will amount to $936 in 5 years, at 6% ? 

By formula, 

A $936 

-= - 06~ 



By 100% method. 

1. 100%=principal. 

2. 6%=interest for 1 year. 

3. 3Q%=5 times 6%=interest for 5 years. 

4. 100%+30%=130%=amount. 

5. $936=amount. 

6. .-. 130%=$936, 

7. 1%=^ of $936=$7.20, and 

8. 100%=100 times $7.20=$720=principal. 

III. .'. $720=the principal that will amount to $936 in 5 years 
at 6%. 



106 FJNKEL'S SOLUTION BOOK. 

I. In what time will any sum quadruple itself at 

1. 100%=principal. Then 

2. 400%=the amount 

3. .-. 400% 100%=300%=interest. 

4. 8%=interest for 1 year. 

5. 300%=interest for 300-=-8, or 37 years. 



III. .-. Any principal will quadruple itself in 37-J years at 



II. TRUE DISCOUNT. 

1. Discount on a debt payable by agreement at some 
future time, is a deduction made for "cash," or present payment; 
and arises from the consideration of the present 'worth of the debt. 

2. Present Worth is that sum of money which, put oa 
interest for the given time and rate, will amount to the debt at 
its maturity. 

3. True Discount is the difference between the present 
worth and the whole debt. 

Since P will amount to A inn years, P may be considered 
equivalent to A due at the end of n years. 

.*. P may be regarded as the present worth of a given future 
sum A. 

P *- 

~~ 



I. Find the present worth of $590, due in 3 years, the rate 

of interest being 6%. 

By formula, 

A $590 

= - a= 



By 100% method. 

1. 100%=present worth. 

2. 6%=mterest on present worth for 1 year. 

3. 18%=3x6%=interest for 3 years. 

TT ;4. 100%+18%=118%=amount, or debt. 
<5. $590=debt. 

6. .-. 118%=$590, 

7. 1%yfg- of $590=$5, and 

8. 100% =100 times $5=$500=present worth. 

HI. .'. $500=present worth of $590 due in 3 years at 



BANK DISCOUNT. 



107 



I. A merchant buys a bill of goods amounting to $2480; he 
can have 4 months credit, or 5% off for cash : if money 
is worth only 10% to him, what will he gain by paying 
cash? 



i ;,! 

) 



(2. 
(3. 
(4. 

(5. 
(6. 

(7. 

(8. 
(9. 



(11.) 



III. .'. He 



100%=present worth of the debt. 
10%=interest on present worth for 1 year. 
3-j.%=interest for 4 months. 
100%+3i%=103i%=amount of present worth, 

which equals the debt, by definition. 
$2480=the debt. 
... 103i%=$2480, 

1 %= ?7 ^ T of $2480=$24, and 
106$ 

100% =100 times $24=$2400=present worth. 

$2480 $2400=$80=true discount. 

100%=$2480. 

1 %= T fo of $2480=$24.80, [count for cash. 

5%=5 times $24.80=$124=trade discount, or dis- 

.'. $124 $80=$44=his gain by paying cash. 

would gain $44 by paying cash. 

(/?. 3d p., p. 258,prob. 10.) 



1. 




Remark. It is clear that $2480 $124,=$2356 would pay for 
the goods cash. If the merchant had this sum of money on hand, 
it would, in 4 months, at 10%, produce $78.53-j- interest. But if 
he pays his debt he will make $124. Hence he will gain $124 
$78.534=$45.46f. 



III. BANK DISCOUNT. 



1. J$anh Discount is simple interest on the face of a 
note, calculated from the day of discount to the day of maturity, 
and paid in advance. 

2. The Proceeds of a note is the amount which remains 
after deducting the discount from the face. 

CASE I. 

( Face of note, ) 
Given < Rate, and > to find the discount and proceeds. 

f Time. S T^ i { Z?=^X^X 

v ' Formulae, j P==: j7_jr) t 



108 FINKEL'S SOLUTION BOOK. 

I. What is the bank discount of $770 for 90 days, at 6% ? 
By formula, 

=$770 X -06 X ^=$1 1 .935. 



By 100% method. 

II. 100%=$770, 
2. l%= T -^of$770=$7.70, and 
3. 6%=6 times $7.70=$46.20=discount for 1 year. 
4. $11.935=^ of $46.20=discount for 93 days. 

III. /. $11.935=bank discount on $770 for 90 days at 6 



CASE II. 
ds, ) 
and > to fi 

( Rate 



C Proceeds, ) 
Given < Time, and > to find the face of the note. 



TT 77 

Formula, F=- - . 
1 rn 

I. For what sum must a note be made, so that when dis- 
counted at a bank, for 90 days, at 6% the proceeds will 
be $393.80? 
Bv formula, 

P $393.80 



By 100% method. 

1. 100%=face of the note. 

2. 6%=discount for one year. 

3. 1^%=^ of 6%=discount for 93 days. 

4. 100% lH%=98<nr%=proceeds. 

5. $393.80=proceeds. 

7. 1%= of $393.80=$4, and 



II. 



5. 100% =100 times $4=$400=face of the note. 
III. /. $400=face of the note. 

CASE III. 

Given rate of bank discount, to find the corresponding rate of 

interest. Formula, rate of /.== . 

1 rn 

I. What is the rate of interest when a 60 day note is dis- 
counted at 8% per annum? 
By formula, 



of 7 == 



ANNUAL INTEREST. 109 



II. 



By 100% method. 
4. 100%=face of note. 

2. 8%=discount for 1 year. 

3. 12^ = e^. of 8%=discount for 63 days. 



4. 100% If %=98f %proceeds. 



5. 98|%=100% of itself. 

6. 1%=- of 100% Jff %, and. 

~ f , o%=8 times $%=8 1 y g -%=rate of interest 
III. .-. The rate of interest on a 60 day note discounted at 8% 
per annum=8^%%. 

CASE IV. 

Given the rate of interest, to find the corresponding rate of 

,, . rate of /. 

discount Formula, 



I. What is the rate of discount on a 60 day note which yields 
10% interest? 

By formula, 



By 100% method. 
. 100%=proceeds. 

2. 10%=interest on proceeds for 1 year. 

3. 11%=^ O f 10%=interest on proceeds for 63 days. 

4. 100%+l|%=101|%=face of note. 

5. 101|%=100% of itself. 



II. 



7. 10%=1 
III. " .-. The rate of discount 

Note. It must be borne in mind that the interest on the pro 
ceeds is equal to the discount on the face of the note. 



IV. ANNUAL INTEREST. 

1. A.in/nAJWJ& Interest is the simple interest of the princi- 
pal and each year's interest from the time of its accruing until 
settlement. 

I. No interest having been paid, find the amount due Sept. 
7, 1877, on a note of $500, dated June 1, 1875, with 
interest at 6%, payable semi -annually. 



no 



FINKEL'S SOLUTION BOOK. 



2 yr. 3 mon. 



1 yr. 9 mon. 6 d. 



1 yr. 3 mon. 6 d. 



9 mon. 6 d. 



1st 



2d 



3m.6d 



3d 



4th 



6d 



II. 



(6.) 



(1.) 100%=$500, 187797 

(2.) 1%= T ^ of $500=$5, and 18756 1 

(3.) 6%=6 times $5=$30=simple interest 23 6 

for 1 year. 
(4.) $68=2 T \ times $30=simple interest for 2 years, 3 

months, 6 days. 
(5.) $15=J of $30=semi-annual interest. 

1. 100%=$15, ' 

2. l%= T foof$15=$.15, and 

3. 6%=6 times $.15=$.90=interest on one semi- 

annual interest for 1 year. 

4. $3.885=4^-f times $.90=interest on one semi-annual 

interest for the sum of the periods each draws int. 
(7.) V. $500+$68+$3.885=$571.885=amount of the note. 
III. V. $571.885=amount of the note. 

Explanation. At the end of six months there is $15 interest 
due ; and, since it was not paid at that time, it drew interest 
from that time to the time of settlement, which is 1 yr. 9 mon. 
6 da. At the end of the next six months, or at the end of the 
first year, there is another $15 due; and, since it was not paid at 
that time, it drew interest from that time to the time of settle- 
ment, which is 1 yr. 3 mon. 6 da. In like manner, the third 
semi-annual interest drew interest for 9 mon. 6 da., and the 
fourth for 3 mon. 6 da. This is the same as one semi-annual 
interest drawing interest for the sum of 1 yr. 9 mon. 6 da., 
1 yr. 3 mon. 6 da., 9 mon. 6 da., 3 mon. 6 da. In the dia- 
gram, the line A B represents 2 yr. 3 mon. 6 day., A 1 repre- 
sents the first year the note run, and 1-2 represents the second 
year the note run. Between A and 1 is a small mark that de- 
notes the semi-annual period ; also one between 1 and 2. By 
such diagrams, the time for which to compute interest on the 
simple interest may be easily found. 

I. The interest of U. S. 4% bonds is payable quarterly in 
gold; granting that the income from them might be 
immediately invested, at 6%, what would the income 
on 20 1000-dollar bonds amount to in 5 years, with gold 
at 105? 




ANNUAL INTEREST. 



Ill 



Syr. 



$200 for 4 yr. 9 mon. at 6^, 



$200 for 4 yr. 6 mon. at 656. 



$200 for 4 yr. 3 mon. at 6$. 



2d 3d I 4th 



&c. 



10 11 12 



16 17 1 



II. 



(i- 

(2. 



{I: 

(v. 



$1000=par value of one bond. 

$20000=par value of 20 bonds. 

100%=$20000, 

1%= T ^ of $20000=$200, and 

4%=4 times $200=$800=income for one year. 

$4000=5 times $800=income for five years. 

$200=i of $800=interest due at the end of first 
quarter, and which draws interest to time of set- 
tlement. 

100%=$200, 

1%=-^ of $200=$2, and [est for one year. 

6%=6 times $2=$12=interest on quarterly inter- 

$570=47-^ times $12=interest on quarterly inter- 
est for the sum of 4f yr.-f-4-^- yr.-f-4^ yr.-j- 

-|- i yr. , or 47-J years. 

/. $4000+$570=$4570=income of bonds in gold. 

$1.00 in gold=$1.05 in currency. [rency. 

$4570 in gold=4570 times $1.05=$4798.50 in cur- 

III. .-. The bonds yield $4798.50 in currency. 





'1. 




2. 


(8.). 


3. 
4. 


(9.) 




(10.) 


1(11.) 



Explanation. It must be borne in mind that the quarterly in- 
terest, $200, is put on interest at 6% as soo'n as it is due. At the 
end of the first quarter there is $200 due which draws interest at 
6% for the remaining time, 4 years, 9 months. The second 
quarterly interest is due at the end of six months and draws in- 
terest for the remaining time, 4 years 3 months, and so on with 
the remaining quarterly payments. This is the same as one 
quarterly payment drawing interest for the sum of 4| yr.-|-4J yr. 
yr.+etc., or 47-^- years. 



I. What was due on a note of $1200, dated January 16, 1883, 
and due Aug. 1, 1892, and bearing interest at 8, 
payable annually, if the 2, 3, 5, and 7th years' 
were paid ? 



interest 



112 



FINKEL'S SOLUTION BOOK. 



$96 for 9 yr. 6 m. 15 d. at 85*. 



$96 for 6 yr. 6 m. 15 d. 



4yr. 6m. 15 d. 



$96, 2yr. 6m. 15 d. 



1 yr.6 m. 15 d 



15 d 



II., 




(8.) 



(9.) 
1(10) 



100%=$1200. 

1%=^ of $1200=$12, and 

8%=8 times $12=$96=simple int. for one year 
$480=5 X$96=five simple interests. 
=J of $96=interest for 6 months. 

$48=interest for 15 days. 
.*. $532=simple interest unpaid. 

1. 100%=$96. 

2. l%= T i ? of $96=$.96, and [simple interest. 

3. 8%=8 times $.96=$7.6S=interest on one year's 

4. $193.92=25i times $7.68=interest on year's sim- 

ple interest for 9 yr. 6 mon. 15 da.,+6 yr. 6 mon. 
15 da.,+4 yr. 6 mon. 15 da.,-f-2yr. 6 mon. 15 da., 
+1 yr. 6 mon. 15 da., -(-6 mon. 15 da., or 25 yr. 
3 mon. 

.'. $532+$193.92=$725.92=amount of interest 
due. [1, 1892. 

$1200+725.92 = $1925.92 = amount due July 



III. .-. $1925.92=whole amount due Aug. 1,1892. 
V. COMPOUND INTEREST. 

1. Compound Interest is interest on a principal formed 
by adding interest to a former principal. 

Let / > =principal on compound interest. 



./?:=( l-j-/')=amount of one dollar for 1 year. 
P (l+r)=/ ) 7?=amount of P dollars for 1 year. 
P (l-|-r) 2 =/ > ^? 2 =amount of P dollars for 2 years. 
P (l+/) 3 ==^? 3 =amount o f P dollars for 3 years. 
P (l-[-r) n =f>l? n =amount of P dollars for n years. 
Let ^4=amount of P dollars in n years, and 

/the compound interest of P dollars for years. 
Then I==P^P ..... I. 
. II. 



.III. 



COMPOUND INTEREST. 113 

n 

.*. JR=\/T IV. Applying logarithms to 

p 

n log. /?=log. A log. P ', whence 
_log. A log. P 

log./? 

When compound interest is payable semi-annually. 
P (1-f-f )=amount of P dollars for 4- year. 
p (!-(-!-) 2=amount of P dollars for"l year. 
P (l-j-f) 2n =amount of P dollars for n years. 
.-. A=P (l-f-J) 2n , when payable semi-annually. 
When compound interest is payable quarterly, 
P (1-f-f )=amount of P dollars for year. 
P (1-j-f ) 2 =amount of P dollars for -J- year. 
P (1-j-f ) 3 =amount of P dollars for year. 
P (1-j-f ) 4 =amount of P dollars for 1 year. 
P (1-j-f ) 4n amount of P dollars for n years. 



When the interest is payable monthly, 
A-(1+A). 

When the interest is payable q times a year, 
A = P (l+|) qn - 

CASE I. 

( Principal, } 
Given < Rate, and > to find the compound interest and amount. 

( Time, ) ( I=PR*P, 

Formula,, 



I. Find the compound interest and amount of $500 for 3 

years at 6%. 
By formulas, 

^=/>7? n =:$500X(l+.06) 3 =$595.508, and 
I=PR* />=$500 X ( 1+.06) 3 $500=$95.508. 
Remark. In compound interest, the 100% method becomes 
very tedious. 

By 100% method. 
' (1.) 100%=$500, 

(3.) 6%=6 times $5= $30=interest for 1 year. 
(4.) $500+$30-=$530=arnount, or principal for the 
second year. 

.. 100%=$530, 

2. 1 %= r ^ r of $530=45.30, [year. 



r IL^ 



(5.) 



3. 6%=6 times $5.30=$31.80=interest for second 

4. $530+$31.80=$561.80=amount, or principal for 

the third year. 



114 



FINKEL'S SOLUTION BOOK. 



III. 



1. 100%=$561.80, 

2. 1 %= T for of $561.80=45.618, and [year. 
(6.) 3. 6%=6 times $5.618 $33.70S=interest for third 

4. 561.80+$33.7.08=$595.508=amount at end of the 

third year. 
(7.) X $595.508 $500=$95.508=compound interest. 

( $95.508=compound interest, and 
'"' ( $595.508=compound amount. 



CASE II. 



C Principal, 
Given < Rate, and 

( Compound Interest, 



to find the time. 
Formula, n= 



log. A log. P 



log./? 

I. In what time will $8000 amount to $12000, at 6% com- 
pound interest? 
By formula, 

log. A log. JP = log. 12000 log. 8000^ 
log. R log. 1.06 

4.0791813.903090 _ 1K .... 

=6 yr. 11 mon. 15 da. We may solve the 

problem thus: $8000( 1.06 ) n =$ 12000, whence (1.06) n =12000-j- 
8000=1.50. Referring to a table of compound amounts and 
passing down the column of 6%, we find this amount between 6 
years and 7 years. 

The amount for 6 years is 1.4185191 ; the amount for required 
time is 1.50. .'. There is a difference of 1.50 1.4185191, or 
.0814809. The difference for the year between 6 and 7 is 
.0851112. .0851112 amount for the whole period between 6 
and 7, .0814809amount for g%tiif of the period or, 11 mon. 15 
da. /. The whole time=6 yr. 11 mon. 15 da. 



CASE III. 

( Principal, ^ 

Given < Compound Intesest or Amount, and > to find the rate. 
( Time, ) 

Formula, r=n l^ 1. 

I. At what rate, by compound interest, will $1000 amount 
to $1593.85 in 8 years? 
By formula, 



ANNUITIES. 115 



CASE IV. 

C Compound Interest or Amount } 

Given ] Time, and V to find the principal 

( Rate, \ 




I. What principal, at compound interest will amount to 
27062.85 in 7 years at 4% ? 
By formula, 



CHAPTER XIV. 

ANNUITIES. 

I. An, Annuity is a sum of money payable at yearly, or 
other regular intervals. 

!1. Perpetual, or 
I: SSKor 
4. Contingent. 

3. A Perpetual Annuity is one that continues forever. 

4. A Limited Annuity ceases at a certain time. 

5. A Certain Annuity begins and ends at fixed times. 

6. A Contingent Annuity begins or ends with the 
happening of a contingent event. 

7. An Immediate Annuity is one that begins at once. 

8. A Deferred Annuity is one that does not begin im- 
mediately. 

9. The Final or Forborne value of an annuity is the 
amount of the whole accumulated debt and interest, at the time 
the annuity ceases. 

10. The Present Value of an annuity is that sum, 
which, put at interest for the given time and given rate, will 
amount to the initial value. 

II. The Initial Vallie of an annuity is the value of a 

deferred annuity at the time it commences. 



116 



FINKEL'S SOLUTION BOOK. 



to find the initial value of a perpetuity. 



CASE I. 
C Annuity, ) 
Given < Time, and 

( Rate, 

I. What is the initial value of a perpetual annuity of $300 a 
year, allowing interest at 6% ? 

1. iOO%=initial value. 

2. 6%=interest for 1 year. 

3. $300=interest for 1 year. 

4. .-. 6%=$300. 

5. \/ c =\ of $300=$50, and 

. 100%=100 times $50=$5000=initial value. 
III. ^ .-. Initial value=$5000. (R* H. A., p. 310, prob. 1.) 

I. What is the initial value of a perpetual leasehold of $2500 
a year payable quarterly, interest payable semi-annually 
at 6%; 6% payable annually ; 6% payable quarterly? 

1. Let 5=the annuity. Then 6'=the amount due in 

3 months. 

2. 6'-)-5'(l-|-^)=namount due in 6 months. 

3. /. v4 = 5+6'(l-f-.01i-)=:$625 + $625(1.01|) = 

$1259.37i=amount due at the end of 6 months. 

4. 100% initial value. 

5. 3%=semi-annual annuity. 

6. $1259.37^=semi-annual annuity. 



HJ 



B.< 



8. l%=Jof $1259.37i=$419.7916f , and 

9. 100%=100 times $419.7916f=initial value. 

1. Let .Sr^amoiint due in 3 months. Then 

2. S-|-5 l (l-f-j)==ainount due in 6 months, [and 

3. 6 1 4-6 1 (l4-T)+^(l+| r )= amount due in 9 months, 

4. S+S(l+-J)+S(l+} r )+S(l+J r )== amount due 
in 1 year. [(1+'V)^ 2 556.25. 

5. .-. ^=$625+$625( l+-\ 6 )+ $625(1+' V 2 )+ $625- 

6. 100%=5nitial value. 

7. 6%=annuitv. 

8. $2556.25=annuity. 

9. .-. Q%= $2556.25. 

10. 1%=4 of $2556.25=$426.0416|, and [value. 

11. 100%=100 times $426.0416f =$42604. 16f=initial 

1. 100%==initial value. 

2. l-^%=quarterly annuity. 

3. $625=quarterly annuity. 



5. 1^= of $625-4416.6666|, and 

6. 100%=100 times $416.6666|=$41666. 




ANNUITIES. 

Initial value of A=$41979.16f, 
Initial value of B=:$42604.16f , 
Initial value of C=$41666.66f . 

(R. 



117 



CASE II. 



Given 



Annuity, 
Interval, 
Rate, and 



and 



H. A., p.S10,prob.5.) 



to find the present 
value of a deferr- 



Time the perpetuity is deferred, J ed perpetuity. 



Let 6*=the annuity, f the rate, and R=l-\-r. Then by Case 
I., the initial value of S is S-~-r. To find the present value of 
the initial value, we use formula III., compound interest. .-. P 

S S 

au ,. , rr= nu D -^-r- in which / is the time the perpetuity 
r (\-\-rJ- R'(R 1) 

is deferred. 

I. Find the present value of a perpetuity of $250 a year, de- 
ferred 8 years, allowing 6% interest. 
By formula, 

$250 _ $250 

~R*(R 1) (l-f-.06) 8 (l+.06 1)~~.06(1.06) 8 ~~ 
By 100% method. 
(1.) 100%=initial value. 
2.) 6%=annuity. 
3.) $250=annuity. 
4.) /. 6%=$250. 
5.) l%=i of $250=$41|, and 
!!.<! (6.) 100%=100 times $41f=$4166.66=mitial value. 

1. 100%=present value of $4166.66| due in 8 years 

at 6%. 

2. 159.38481 %==( 1.06) 8 XlOO%=compound amount 

of the present value for 8 yr. at 6%. 

3. /. 159.38481 %=$4166.66f, 

4. l^^i-^Wr of $4166.66f=$26. 1422, and 



III. 



I. 



5. 100%=100 times $26.1422 = $2614.22 = present 

value. 

/. The present value of a perpetuity of $250 a year He- 
ferred 8 years at 6% interest=$2614.22. 

Find the present value of an estate which, in 5 years, is 
to pay $325 a year forever; interest 8%, payable semi- 
annually. 

By formula, 

5 $325 $325 



"[ 
$2 



.0816(1.04) 10 



690.67. 



118 FINKEL'S SOLUTION BOOK. 

By 100% method. 
(1.) 100% initial value. 
(2.) 4% amount due in 6 months. 
(3.) 4%+(1.04)x4% 8.16% amount due in 1 year. 
(4.) $325 amount due in 1 year. 
(5.) .-. 8.16% $325, 

(6.) 1%=^ of $325 $39.828431, and [value. 

(7.) 100% 100 times $39.828431= $3982.8431 = initial 

1. 100% present value of $3982.8431. 

2. 148.024428f*:=(1.04) 10 X100%==compound amount 

of 100% for 5 yr. at 8%. 



II. 



1(8.) 



3. .-.148.024428% $3982.8431, 



4. l%=T48.<ji44?8 f $3982.8431 $26.9067, and 

5. 100% 100 times $26.9067 $2690.67 present 

value. 
III. .'. $2690.67 present value of the estate. 

(JR. H. A., p. 311,prob. 4.) 

Explanation. The initial value is a sum of money which 
placed on interest at 8% payable semi-annually will produce 
$325 per year. But 8% payable semi-annually is the same as 
8.16% payable annually. Hence 8.16% is the annual payment. 
But $325 is the annual payment. Hence 8.16% $325, from 
which we find that $3982.8431 is the initial value, or the amount 
that will produce $325 per year. Then the present value of a 
sum of money that will pay $325 is $3982.8431 if the payments 
are to begin at once, but $3982.8431-r- (1.04) 10 if the payments 
are not to begin until the end of 5 years. 

CASE III. 



Given -< . nnul , ty> , Uo find the present valve of an an- 

I lime to run, and I ., ~. . 

T . T nuity certain. 

(^Interval, ) 

(a) Let P denote the present value. The amount of P for n 



Let 6* the payment, or amount due the first year. 
= the amount due the second year. 

- the amount due the third year. 

S'-|-S 1 /?--S7? 2 +S7? 8 ==the amount due the fourth 
year. [due the nth year. 

S+ Sfi+Sfi'*+ Sfi*+ ....... + 67?"- 1 amount 

.-. A= 



...(1) 
AR= SR 



(2), by multiplying (1) by R. 
ARA=SR*S. . . (3), by subtracting (1) from (2). 

1) 4. But PR*=A. 




ANNUITIES. 119 



'" ( 5 ')> whence 

- 

- 



When the annuity is to begin at a certain time, and then 
to continue a certain time. 

Let/ the number of years the annuity is deferred, and q= 
the number of years the annuity continues. Then 

, S ~Rp+q 1 

P=-= - X T?p+q =the present value of an annuity S, for 

the time (p-\-y) years, and 

C Dp 1 

P"= -X D^ =t h e present value of an annuity S, for p 




R P+V ) R 1 R P+V 

I. Find the present value of an annuity of $250, payable an- 
nually for 30 years at 5%. 
Given S, n, and r. 
By formula, 

S R-\ $250 (1.05)so-l ,0040110* 
^7?=r X ~7F~ r05~ X (1.05)30 : 
By 100% method. 
(1.) 100% initial value. 
(2.) 5% annuity. 
(3.) $250 annuity. 
(4.) .-. 5% $250, 
(5.) l%-=i of $250$50, and 
(6.) 100% 100 times $50%=$5000=initial value of 

an immediate perpetuity of $250 per year. 
'1. 100% present value of an annuity deferred 30 
years. [ent value for 30 years. 

2. 432.19424% (1.05) 30 XlOO% amount of pres- 

3. .-. 432.19424% $5000, 

4. l%= T7 ^ T rre of $5000 $11.568865, and 

5. 100%=100 times $11.568865 $1156.8865 pres- 

ent value of annuity of $250 deferred 30 years. 
(8-) " /. $5000 $1156.8865=$3843.1135 present value 

of an annuity continuing 30 years. 
.!II. V. $3843.1135 present value of an annuity of $250, payable 
annually for 30 years, 



i20 



FINKEL'S SOLUTION BOOK. 



Remark. Since $5000 is the initial value which, in this case, 
is also the present value of an immediate perpetual annuity, or 
perpetuity of $250, and $1156.8865 the present value of an an- 
nuity of $250 deferred 30 years, $5000 $1156.8865 $3843.1135= 
the present value of an annuity of $250 continuing for 30 years 



I. 



II. 



Find the present value of an annuity of $826.50, to com- 
mence in 3 years and run 13 years, 9 months, interest 
6%, payable semi-annually. 

Given =$826.50, r=.06, /=3 years, and ^=13| years. 
When interest is payable semi-annually, _/?=(l-[-|-) 2 . 
By formula (7), 

$826.50.. a0609)^ 1=$6324m 



' .0609 ' (1.0609) 16 ^ 



(8.) 



(9.) 



(10.) 



III. 



By 100% method. 
(1.) 100%=initial value. 
(2.) 3%=amount due in 6 months. 
(3.) 3%+3% (1.03)=6.09%=amount due in 1 year. 
(4.) $826.5Q=amount due in 1 year. 
(5.) .-. 6.09%=$826.50, 
(6.) l%=Tfo of $826.50=$135.712643, and 
(7.) 100% = 100 times $135.712643 = $13571.2643= 
initial value 

1. 100%=present value of a perpetuity of $826.50 

deferred 3 years. 

2. 119.40523 %=( 1.0609 ) 2 times 100%=amount of 

present value for 3 years. 

3. .-. 11940523%=$13571.2643, 

4. l%=mr.ArB7Tr f $13571.2643=$113.6686, 

5. 100%=100 times $113.6586=$11365.86=present 

value of such a perpetuity deferred 3 years 

1. 100%=present value of such a perpetuity deferr- 

ed 16f years. 

2. 269.212027%=(1.0609) 16 ^ times 100% = amount 

of present value for 16f years 

3. /. 269.212027%=$13571.2643, . 

4. 1%=^^^ of $13571.2643=150.4117, 

5. 100%=100 times $50.4117= $5041.17= present 

value of such a perpetuity deferred 16f years. 

.-. $11365.86 $5041. 17=$6324.69=present value 
of an annuity of $826.50 deferred 3 years and 
continuing 13| years. 

$6324.69=present value of $826.50, etc. 



If the 



ANNUITIES. 



121 



If the annuity is to begin in p years and continue forever, the 
formula, 

S 



if ^=00, the 



For, since P= 
quantity 



=1 =1 0, approaches 1 as its limit, 

CO 



and we have ^ 






I. Find the present value of a perpetual annuity of $1000 to 
begin in 3 years, at 4% interest. 

By formula, [value of the annuity. 

8 $1000 



II, 



By 100% method. 
(1.) 1 00 %=initial value. 
(2.) 4%=annuity. 
.) $1000=annuity. 



3 

(4-) 

(5.) l%=i of $1000=$250, and [$1000. 

(6.) 100%=100 times $250=$25000=initial value of 

1. 100%=present value. 

2. 112.4864% = (1.04) 3 times 100% = amount of 

present value for 3 years at 4%. 

3. .-. 112.4864 %=$25000, 

4. l%=nnriimr of $25000=$222.2492, and 

5. 100%=100 times $222.2492=$22224.92=present 

value. 

III. .-. $22224.92=present value of an annuity of $1000 to be- 
gin in 3 years at 4%. 



f Annuity, 
J Rate, 
\ Interval, and 



CASE IV. 
-to find the final or forborne value. 



Given 

' 

(^Tirne to run, 
Let $=amount due first year. 

$_|_ l $ f j ff=amount due second year. 

nt due third year. 

=:amoun t due the fourth year. 

2 + & ff3 + +o'7? n - 1 = amount due 

the nth year. 
Let ^4=amount due the ^th year. 

- 1 ... (1). 



122 FINKEL'S SOLUTION BOOK. 



. . (2), by multiplying (1) by R. [from (2). 

AR A=Sfi" S ...... (3), by subtracting (1) 



A pays $25 a year for tobacco ; how much better off would 
he have been in 40 years if he had invested it at 10% 
per annum? 

By formula, 



1] = $11064.8139. 



II. 



By 100% method. 

1. 100%=initial value. 

2. 10%=annuity. 

3. $25=annuity. 

4. .-. 10%=$25, 

5. 1%= T V of$25=$2.50, and 



6. 100%=109 times $2.50=$250=initial value. 

7. $44.2592556=[(1.10) 40 1]X$1 compound interest of 

$1 for 40 yr. at 10%. [$250 for 40 yr. at 10%. 

I 8. .'. $11064.8139=44.2592556 X$250=compound int. of 
III. /. He would be $11064.8139 better off. 

Remark. $250 placed on interest at 10% will produce $25 
per year. If this interest be put on interest at 10%, instead of 
spending it for tobacco, it will amount to $11064.8139 in 40 
years. This would be a very sensible and profitable investment 
for every young man to make, who is a slave to the pernicious 
habit. 

I. An annuity, at simple interest 6%, in 14 years, amounted 
to $116.76 ; what would have been the difference, had it 
been at compound interest 6% ? 
(1.) 100%=initial value, or the principal that would 

produce the annuity. 
(2.) 6%=annuity for 1 year. 
(3.) 84%=14x6%=annuity for 14 years. 

1. 100%=6%, 

2. 1%= T ^ of 6%=-^Q-%, and [1 year. 
(4.) 3. 6%=6 times -^ r %=2T%=interest on annuity for 

4. 32.76 %=91 times 2T%=interest on annuity for 

(1+2+3+ +14), or 91 years. 

(5.) 84%+32.76%=116.76%=whole amount of the 
n.< annuity. 

(6.) $116.76=whole amount of the annuity. 

(7.) /. 116.76%=$116.76, 

(8.) 1%=^^ of $116,76=$!, and 

(9.) 100%=100 times $l=$100=initial value. 

(10.) 6%=6 times $l=$6=annuity. 



ANNUITIES. 



122 



(11.) $1.260904 [(1.06) I* l]x$l= compound inter- 
est on $1 for 14 yrs. at 6%. 

(12.) $126.0904 1.260904 X $100 compound interest 
on $100 for 14 yrs. tit 6%. 

(13.) .-. $126.0904 $116.76 $9.3304 difference. 
III. .-. The difference=|9.8304. 



CASE V. 
Final Value or Present Value 



Given < Rate, and 

( Time to run, 

c< 



Solving ^ 



r> n _ -J 



to find the annuity. 



R 1 



with respect to 6* and we have 
T (1). If ^l=the final or 



forborne value, by the formula in the last case, we have A- 
~ 1. Solving this with respect to 6", we have. 



S=- 



(2). 



I. How much a year should I pay, to secure $15000 at the 

end of 17 years, interest 7% ? 
By formula (2), 

rA .07 X $15000 
o Ou -= = $4ob.o& 

By 100% method. 

(1.) 100% annuity. 

(2.) 7% annuity. 

(3.) .-. 7% 100%, 

(4.) l%=y of 100% 14f%, and 

(5.) 100% 100 times 14f% 1428f % initial value. 

1. 100% present value of 1428^% due in 17 years. 

2. 315.8815% amount of present value for 17 years. 

4. i%=-j fTT .fo TT of 142 T 8|%=4.522591%, and 

5. 100% 100 times 4.522591% 452.2591% pres- 
II. < ent value. 

(7.) .'. 1428^% 452.2591% 976.3223% present 

value of an annuity running 17 years. 
(8.) 3.1588152% (1.07) 17 times l%=amount of 1% 

for 17 years. 
(9.) 3084.0217% (1.07) 17 times 976.3223 %=amount 

of 976.3223% for 17 years at 7%. 
(10. $15000 amount, or final value. 

(11. .- 3084.0217% $15000. 
(12. l%=inrreWTT of $15000 $4.8638, and 
*( 13. 100%=100 X $4.8638 $486.38 annuity. 



124 FINKEL'S SOLUTION BOOK. 

III. .-. I must pay $486.38. 

CASE VI. 

( Annuity, ) , ~ , ,. .. 

s^. i T i TT i c J.T- A *. j i ^o nna time it 

Given < Present Value of the Annuity, and } 

) T- * \ runs. 

( Rate, ) 

In formula (6), Case III., we have P= -x ^ , whence 

Pr_SPr 

S~ ~~S~* 

..^? n = ^ (1). Applying logarithms, 




S \ log. 5 log. (SPr) 

-- - ~ 



I. In how many years can a debt of $1,000,000, drawing 
interest at 6%, be discharged by a sinking fund of 
$80, 000 per year? 
By formula (2), 
^ A o g . s log. ( S Pr ) == \og. 80000 log. ( 800001000000 X .06 ) 

log. R log. 1.06 

log. 80000 log. 20000^4.903090 4.301030_.6Q2060 := 
log. 1.06 .025306 ~ 025306" 

years. 

By another method. 

Assume $1,000,000 to be the present value of an annuity of 
$80000 a year. Then $12.50 may be considered as the present 
value of $1 for the 'same time and rate. By reference to a table 
of present worths $12.50, which is 1000000-f-80000, will be 
found to be between 23 and 24 years. 

Note. A table of present worths may be computed by form- 
ula (6.), Case III., in which put 6'=$1. 

I. In what time will a debt of $10000, drawing interest at 
6%, be paid by installments of $1000 a year. 
By formula, 

log. S log. (S 7V)_log. 1000 log. ( 100010000 X. 06) 
log. R log. 1.06 

32.602060 1KI70 - 1K 

-=15.725 years=15 yr. 8 mo. 21 da. 



.025306 

By another method. 

Assume $10000 to be the present value of an annuity of $1000 
a year. Then $10000-r-1000=$10=the present value of $1 for 
the same time and rate. By referring to a table of present worth 
we find this amount between 15 and 16 years. .*. The time is 15 
years -f- 



ANNUITIES. 125 

The compound amount of $10000 for 15 yr. at Q%= $23965.58 
The final value of $1000 for 15 years at 6%= $23275.97 

Balance^ $ 689.61 

This balance, $689.61, will require a fraction of a year to dis- 
charge it. The part of a year required, will be such a fraction of 
a year as the amount of $689.61 for \\\e fraction of a year is of 
$1000. 

6% of $689.61 for fae fraction of a year=$41.3766X fraction 
of a year. 

.-. $689.61+$41.3766X fraction of a year=the amount of 
$689.61 for the fraction of a year. This amount divided by 
$1000, a yearly payment, will give infraction. 
$689.61+141.3766 ^fraction 

-$1655~ =/'**>* whence 

$689.61+141.3766 X fraction=$\(yM X fraction 

.. $1000 yJraction$\. 3766 X/c#o==$689.61, or 



689.61 

=& months, 19 days. 



.. The whole time=15 yr. 8 mon. 19 da. 

CASE VII. 

C Annuity, . } 

Given < Time to Run, and > to find the rate of in- 

( Present Value of an Annuity, ) terest. 

From the formula (6), Case III, *=-& ?X p n we ODta i n 

=-^r .... (1). This is the simplest expression we can ob- 
tain for the rate as the equation is of the ^th degree and can not 
be solved in a general manner. 

I. If an immediate annuity of $80, running 14 yr., sells for 
$650, what is the rate? 

By formula, 

^ n 1_^__$650 _ or 



^-=8.125. Solving this equation by the method of 

Double Position, we find r=S%-\-- 

By another method. 

$65Or-$80=8.125. By referring to a table of present worths 
of $1, corresponding to 14 years, we find it to be between 8 and 



126 FINKEL'S SOLUTION BOOK. 



PROBLEMS. 

1. What is the amount of an annuity of $1000, forborne 
15 years, at 3% compound interest? Ans. $19295.125 



2. What will an annuity of $30 payable semi-annually, 
amount to, in arrears 3 years at 7% compound interest? 

Ans. - 

3. What is the present worth of an annuity of $500 to con- 
tinue 40 years at 7% ? Ans. - 

4. What is the present worth of an annuity of $200, for 7 
years, at 5% ? Ans. $1152.27. 

5. A father presents to his daughter, for 8 years, a rental of 
$70 per annum, payable yearly, and the reversion for 12 years 
succeeding to- his son. What is the present value of the gift to 
his son, allowing 4% compound interest? Ans. - 

6. A yearly pension which has been forborne for 6 years, at 
6%, amounts to $279 ; what was the pension? Ans. $480.03. 

7. A perpetual annuity of $100 a year is sold for $2000 ; at 
what rate is the interest reckoned? Ans. - 

8. A perpetual annuity of $1000 beginning at the end of 10 
years, is to be purchased. If interest is reckoned at 3-J%, what 
should be paid for it? Ans. - 

9. If a clergyman's salary of $700 per annum is 6 years in ar- 
rears, how much is due, allowing compound interest at 6% ? 

Ans. $4882.72. 

10. A soldier's pension of $350 per annum is 5 years in ar- 
rears; allowing 5% compound interest, what is due him? 

Ans. $1933.97. 

11. What annual payment will meet principal and interest of 
a debt of $2000 due in 4 year a 8% compound interest? Ans. 

12. What is the present worth of a perpetual annuity of $600 
at 6% per annum? Ans. $10000. 

13. What is the present value of an annuity of $1000, to com- 
mence at the end of 15 years, and continue forever, at 6% per 
annum? Ans. $6954.40. 

14. To what sum will an annuity of $120 for 20 years amount 
at 6% per annum? Ans. $4414.27. 

15. A debt of $8000 at 6% compound interest, is discharged 
by eight equal annual installments; what was the annual install- 
ment? Ans. $1288.286- 



MISCELLANEOUS PROBLEMS. 



127 



CHAPTER XV. 

MISCELLANEOUS PROBLEMS, 

INVOLVING THE VARIOUS APPLICATIONS OF PERCENTAGE. 

I. Sold a cow for $25, losing 16f% ; bought another and sold 



it at a gain of 16% ; I neither gained nor lost on the two ; what 
Was the cost of each? 

-1. 100% cost of the first cow. 

2. 16f % loss. 

3. 100% 16|% 83^% selling price. 

4. $25 selling price. 

5. .-. 83-J%- $25, 



A. 



B. 



=- of $25 $.30, and 



in. 



6. 



7. 100% 100 times $.30 $30=cost of first cow. 

8. $30 $25 $5, loss on the first cow, and gain on 

second cow. 

1. 100% cost of second cow. 

2. 16% gain. 

3. $5 gain. 

4. .-. 16% $5. 

5. 1% T V of $5 $.3125, and [cow. 

6. 100% 100 times $.3125 $31.25 cost of second 
( $30=cost of first cow, and 

) $31.25=cost of second cow. 



Remark. Since I lost $5 on the first cow, and neither gained 
**r lost on the two, I must have gained $5 on the second cow. 
. 16% $5. 

I. There have been two equal annual payments on a 6% note 
of $175, given 2 years ago this day The balance is 
L40 ; what was each payment? 



II. 



(1.) 100 %=a payment. 
(2.) 100% $175, 
(3.) l%=ri(5- of $175=$1.75, and 
(4.) 6%= 6 times $1.75=$10.50=interest for 1 year. 
(5.) $175+$10.50= $185.50 amount before paying 
the payment. [payment. 

(6.) $185.50 100% amount left after paying the 

1. 100% $185.50 100%, 

2. 1% T io of ($185.50 100%) $1.855 1%, and 

3. 6% 6 times ($1.855 6%) $11.13 6% inter- 
(7.)J est for second year. 

4. $185.50 100%+$11.136% $196.63 106% = 

amount before paying the last payment. 

5. $196.63 106% 100% $196.63 206% = 

amount left after paying the last payment. 



128 



FINKEL'S SOLUTION BOOK. 



(8. 

(9. 
(10.) 
(11.) 
(12.) 



$154.40=amount after paying the last payment 

.-. $154.40=$196.63 206%. 

206% $196.63 $154.40 $42.23, 

1% ^ of $42.23=$.205, and 

100% 100 times $.205 $20.50=the payment. 



III. .*. |20.50=the payment. 

Remark. In this solution we are obliged to use the minus 
sign, , which is no obstacle to the student of algebra, but to the 
student of arithmetic it may seem insurmountable. To avoid 
this sign, we give another solution. 



II. 



III. 



(I-') 

(2.) 

(3.) 
(4-) 
(5.) 
(6.) 

(7.) 
(8.) 



(9.) 



(10,) 



(11.) 
(12.) 
(13.) 
(14.) 
(15.) 



100% the payment. Then 

$154.40+100% amount of the debt at the end of 

of the second year. 

100% principal that produced this amount. 
6 % interest. 
106% amount. 

... 106% $154.40+100%, [and 

l%- T i of ($154.40+100% )-$1.4566^+M%, 
100% 100 times ($1.4566^+11% ) = $145.66^ 

-|-94i|% amount at end of the first year after 

paying off the payment. 
$145.66^+94/3 %+100% $145.66^ + 194|f % 

amount before paying oft' the payment 

amount at end of first year. 
100% the principal that produced it. 
6%=interest. 
106% amount. 



l-83^fr%, and 
100% = 100 times ($1.3?Hw + 

$137JtfH-183^/V%=tibe amount at first. 
$175=the amount at first. 



and 
100%=100 times $.205 $20.50=the payment. 



.'. $20.50 the payment. 



(R. H. A., p. 26^ prob. 5.) 



Explanation. $154.40=the amount after paying off the last 
payment. .-. $154.40-)-100%=amount before paying of the last 
payment, or it equals the debt at the end of the first year plus 
the interest on this debt for the second year. .-. We let 100%= 
the debt at the end of the first year, 106% amount of 100% for 
1 year. /, 106% = $154.40 + 100%. Then proceed as in the 
solution. 

I. If a merchant sells f of an article for what J of it cost, 
what is his gain % ? 




MISCELLANEOUS PROBLEMS. 



129 



1. 100%=cost of whole article. 

2. 87i%=J of 100%=cost of of the article. 

3. 87^-%=selling price of f of the article. 

4. 29i%= of 87|%=selling price of of the article. 

5. 116f%=4 times 29^%= selling price of the whole 

article. 

6. .'. 116|% 100%=16f%=gain. 

.-. 16|%=his gain. (Milne's Prac., p. 360,prob. 51.) 

A merchant sold goods to a certain amount, on. a commis- 
sion of 4%, and having remitted the net proceeds to the 
owner, received ^% for. prompt payment, which 
amounted to $15.60. What was his commission? 

100%=cost of goods. 

4%=commission. 

100% 4%=96%=net proceeds. 

;!% amount received for prompt payment. 

2. $15.60=amount received for prompt payment. 

3. .-. i%=$15.60. 

4. 1%=4 times $15.60=$62.40. 

5. 100% =100 times $62.40=$6240=net proceeds. 
.-. 96%=$6240. 

1%=^ of $6240=$65, and 

100% =100 times $65=$6500=cost of goods. 

100% =$6500. 

1%=^ of $6500=$65, and 

4%=4 times $65=$260=his commission. 

His commission=$260. 

( Greenleafs N. A., p. J^l.prob. 11.) 



(2.) 
(3.) 



(4.) 



(5.) 
(6.) 
(7-) 

(8.) 



If I sell 30 yards of cloth for $132, and gain 10%, how 
ought I to sell it a yard to lose 25% ? 



$132=selling price of 30 yards. 
$4.40=$132-^30=selling price of one yard. 
100 % cost of one yard. 
10%=gain. 

100%+10%=110%=selling price per yard. 
$4.40=selling price per yard. 
.-. 110%=$4.40. 
1 %= T H of $4.40=$.04, 
100%=100 times $.04=$4=cost per yard. 
I. 100%=$4. 



HI. 



(1.) 

(2.) 

( 3. ) 

(4.) 

(5.1 

(6 ) 

(7. 

(8. 

(9.) 



3. 25% =25 times $.04=$l=loss. 

4. ... $4 ^i^^selling price per yard to lose 25%. 

.. I must sell it at $3 per yard to lose 25%. 

(StoddarcCs Complete, p. 206, prot>. 9.) 




130 FINKEL'S SOLUTION BOOK. 

I. A merchant receives on commission three kinds of flour ; 
from A he receives 20 barrels, from B 25 barrels, and 
from C 40 barrels. He finds that A's flour is 10% better 
than B's, and that B's is 20% better than C's. He sells 
the whole at $6 per barrel. What in justice should 
each man receive? 

(1.) $6=selling price of 1 barrel. 

2.) $510=sellmg price of (20+25+40), or 85 barrels. 

3.) 100%value of C's flour per barrel. 

(4.) 120%value of B's flour per barrel. 

(1. 100%=120%. 

(5.W2. 1%=^ ofl20%=li%, 

13. 10% 10 times 1-J %=12%. 

(6. ) 120% +12 % 132 %=valtie of A's flour per barrel. 



II. 



(7.) 4000%=40 times 100% what C received. 



i! 



(8.) 3000% 25 times 120% what B received. 



2640% 20 times 132% what A received. 
9640% 4000%+3000%+2640% what all rec'd. 
$5 10= what all received. 
9640^=4510. 



(9. 

(10. 
(11. 
(12. 

(13.) 1%=^ of $510 $.52|-J|, and [received. 

(14.) 4000% 4000 times $.52fii $21H||=what C 

(15.) 3000% 3000 times $. 52-Jii $158^1 what B 

received. [received. 

(16.) 2640% 2640 times $.52fi| $139iJ| what A 



v Tici/2"^y A's share, 
ill. .'. I $158.Hf B's share, and 
:C's share. 

( Greentcafs National Aritk. p. 442.) 

I. f of B's money equals A's money. What % is A's money 
less than B's, and what % is B's money more than A's? 

1. 100% B's money. 

2. 75% | of 100% A's money. 

TT j 3. 100% 75% 25%=excess of B's money over A's. 
11X 4. 75% 100% of itself, 

5. 1% T V of 100% H%, and [than A's. 

6. 25% 25 times l%=33i%:=the % B's money is more 

A's money is 25% less than B's, and 

B's money is 33% more than A's money. 

(Stod. Comp.,p. 203,prob. 19.) 

I. At what price must an article which cost 30 cents be 
marked, to allow a discount of 12^% and yield a net 
profit of 16f% ? 



MISCELLANEOUS PROBLEMS. 



131 



II. 



1. 

2. 
3. 

(4. 



100%=30/, 

16f%=16 times ^Xxm5/=profit. 

30/+5/=35/=the price at which it must sell to 



III. 



I. 



1. 100%=marked price. 

2. 12^%=discount from marked price. 

3. 100% 12|%=87i%=selling price. 

4. 35/=selling price. 
(5.n 5. .-. 87|%=35/. 

6. l%=^~ of 35/=.40/, and 

7. 100%=100 times .40/=40/=marked price. 
/. 40^=marked price. 

(Seymour's Prac., p. 203; prob. 4.) 

Had an article cost 10% less, the number of % gain would 
have been 15% more ; what was the gain? 

1. .7#6>%=selling price. 

2. 100% actual cost price. 

3. 100% 100%=gain. 

4. 100% 10%=90%=supposed cost. 

5. 100% 90%=conditional gain. 

6. 90%=100% of itself. 



[difference. 



8. 100% 90%=( 

=conditional gain %. 

9. /. V X100% 100% (100% 

10. 15%=difference. 

11. /. ix^#%=15%. [the actual cost. 

12. 100%=9 times 15%=135%=selling price in terms of 

13. .-. 135% 100%=35%=gain. 

.*. 35%=gain. (R. H. A., p. 406,prob. 87.) 

A literal solution. 
Let 5=:selling price and C=the cost. Then 5 C=gain and 

C* S~* C 1 

-~r- =rate of gain. S T 9 ^C=conditional gain and 

^ 10 $ Q $ Q 

-=conditional rate of gain. .. ^___^.^ or 

.'. 1.35CC=.35C=gain. 



III. 



whence 5=fJ C=1.35 C. 
.-. Rate of gain=.35C-7-C==.35=35%. 

In the erection of my house I paid three times as much for 
material as for labor. Had I paid 6% more for labor, 
and 10% more for material, my house would have cost 
$3052. What did it cost me? 



132 FINKEL'S SOLUTION BOOK. 

(1.) 100%=cost of labor. 

(2.) 300% 3 times 100%=cost of material. 

rl. 100%=100%, i 

, J2. !%=!%, and 

< 6 -')3. 6%=6%. 

U. 100%+6%=106%=supposed cost of labor. 

II. 100% =300%, 

2. l%= T i-o- of300%=3%, and 

3. 10%=10 times 3%=30%. 

4. 300%+30%=330%=supposed cost of material. 

(5.) 330%+ 106 %== 4 36%= :SU PP osed cost of house. 

(6.) $3052=supposed cost of house. 

(7.) .-.436% =$3052, 

(8.) l%=^k of $3052=$7, and 

(9.) 100%=100 times $7=$700=cost of labor. 

(10.) 300%=300 times $7=$2100=cost of material. 

( 11.) $2100+$700=$2800=cost of house. 

III. .-. $2800=cost of the house. 



I. I invest |- as much in 8% canal stock at 104%, as in 6% 
gas stock at 117% ; if my income from both is $1200, 
how much did I pay for each, and what was the income 
from each ? 

(1.) 100%=investment in gas stock. Then 

(2.) 66f %=investment in canal stock. 

1. 100%=par value of the gas stock. 

2. 117%=market value of the gas stock. 

3. .-. 117%=100%,from (1), 

4. 1%=^-^ of 100%=-fr5-% > and 

5. 100%=100 times }fo%=85||%=par value in 

terms of the investment. 
;i. I00%=85|f%, 
(5.)<^2. l%=ff-?-%, and 

U. 6%=6 times |^%=5A%=income of gas stock, 

II. 100%=par value of canal stock. 
2. 104%=market value. 
3. /. 104%=66|%, 
4. l%= T ^of66|%=||%,and 
100%=100 times 



(3.) 



(6.K2. l%=z T ^ of 64^%=|f%, and 

13. 8%=8 times ff %=5^%==mcome o f canal stock. 
(7.) 5^g-%-}-5^=10^%=income from both. 
(8.) $1200=income from both. 
(9.) ,. 



(10.) 1%=-- of $1200=$117, and 



MISCELLANEOUS PROBLEMS. 133 

(11.) 100%=100 times $117=$11700=investment in 

gas stock. [canal stock. 

(12.) 66f%=66f times $117=$7800 = investment in 

(13.) 5A%=5A times $117=$600=income from each. 



$600=income from each. 

JJ f .'. $11700=investment in gas stock, and 
$7800=investment in canal stock. 

I, A man bought two horses for $300; he sold them for $250 
apiece. The gain on one was 5% more than on the 
. other; what was the gain on each? 

1. $300=cost of both. 

2. $500=$250+$250=selling price of both. 

3. $500 $300=$200=gain on both. 

4. 100%=gain on first horse. Then 

5. 105%=gain on second horse. 

n <| 6. 205%=100%+105%=gain on both. 

7. $200=gain on both. 

8. .-. 205%=$200. 

9. l%=?fa of $200=$Jf, and 

10. 100%=100 times $|f=$97.56 4 T =gain on the first. 

11. 105%=105 times $=$102.43f=gain on the second. 

Ill $ $97.56^ T =gain on the first, and 
' ( $102.43f|=gain on the second. 

Note. In this solution, it is assumed that the gain on one was 
5% of the gain on the other more than the other, and this is the 
usual assumption. But the problem really means that the per 
cent, of gain on one, computed on its cost, was 5% more than 
the per cent, of gain on the other, computed on its cost. By this 
assumption, the problem is algebraic. The following is the 
solution: Let #=the cost of the first horse, and $300 #, the 
cost of the second. Then $250 # gain on first, and $250 
($300 x)=x $50, the gain on the second. ($250 x) -t-x= 
rate of gain on the first, and (x $50)-f-($300 x), the rate of 
gain on the second. Then (250 x)-$-x (x 50)-r-(300 x)= 
j 1 ^. Whence, by clearing of fractions, transposing and, combin- 
ing, *2__ 10300 * = i500000, ^=5150^50^10009= $147.7755, 
the cost of the first horse. $300 *=$152.2245, the cost of the 
second horse. $250 #=$102.2245, gain on the first horse, and 
* $50=$97.7755, the gain on the second horse. 

I. An agent sells produce at 2% commission, invests the 
proceeds in flour at 3% commission; his whole commis- 
sion was $75. How many barrels of flour did he buy 
at $5 per barrel ? 



134 



FINKEL'S SOLUTION BOOK. 



II. 



III. 
I. 



II 



(1.) 100%=value of the produce. 

(2.) 2 % the commission. [vested in the flour, 

(3.) 100% 2%=98%=net proceeds, or amount in- 

1. 100%=cost of the flour. 

2. 3%=commission on flour. 

3. 100%+3%=103%=whole cost of the flour. 



(4.) 



(5.) 
(6.) 
(7.) 
(8.) 
(9.) 
(10.) 

(11.) 
(12.) 



.-. 103%=, 
l%=rta of 98%=!%%, and 
lOO%==100X T 9 -o 8 *% = 95^% = cost of flour in 
terms of the value of the produce. 
95 T V\%=2 T 8 o%%=commission on flour. 

%==whole commission. 
$75=whole commission. 



l%^$75-M T 8 oV= $15.45, and [produce. 

100%=100 times $15.45=$1545 = value of the 
95 T V%%==95 T VV times $15.45= $1470 = value of 

the the flour. 
$5=costofl barrel. 
$1470=cost of 1470-=-5, 01 294 barrels. 



\ 
2. 

(3.) 
(4.) 



.*. The agent bought 294 barrels of flour. 

A distiller sold his whisky, losing 4% ; keeping $18 of 
the proceeds, he gave the remainder to an agent to buy 
rye at 8% commission; he lost in all $32 ; what was the 
whisky worth? 

(1.) 100%=value of the whisky. 
>.) 4%=loss. 

100% 4%=96%=amount he had left. 
96% $18=amount he invested in rye. 

1. 100%=cost of the rye. 

2. 8%=commission on the rye. 

3. 100% +8 %=108%=whole cost of rye. 

4. .-. 108%=96% $18, 

5. 1%=^ of (96% $18)=|% $.16|, and 

6. 100%=100 times (f% $.16f)= 88f% $16.66f 

=cost of rye. 

7. 8%=8 times (|% $.16f)=7i% $1.33i=com- 

mission on rve. 

$1.33^-)=! H% $1.33|=wholeloss. 
2=\vhole loss. 
H^%_$1.33^ =: $32 
*o= $33.33i, 

of $33.33ih=$3, and 



(5.) 



(6.) 

(7.) 
(8.) 
(9.) 

(10.) 
l(ll-) 



100%-=100 times $3=$300=value of the whisky. 



III. .-. $300=value of the whisky. 



(Jt. H. A., p. 4Q6, prob 91.) 



MISCELLANEOUS PROBLEMS. 135 

I. What will be the cost in New Orleans of a draft on New 
York, payable 60 days after sight, for $5000, exchange 
being at \\/o premium? 

1. 100%=face of the draft. 

2. l|%=premium. 

3. 100%+1-J%= 101i%=rate of exchange. 

4. 5%=discount for one year. 



II. 



of 5%=discount for 63 days. 



/. 101-4-% f %=100|%=cost of the draft 

7. 100% =$5000. 

8. 1%=^ of $5000=$50, and 

lOOf %=100f times $50=$5031.25=cost of the draft. 

III. .-. $5031.25=cost of the draft. 

Explanation. Observe that since the draft is not to be paid in 
New York for 63 days, the banker in New Orleans, who has the 
use of the money for that time allows the drawer discount on the 
face of the draft for that time, which goes (1) towards reducing 
the premium if there be any, and (2) towards reducing the face 
of the draft. 

Note. The rate of exchange between two places or countries 
depends upon the course of trade. Suppose the trade between 
New York and New Orleans is such that New York owes New 
Orleans $10,250,000 and New Orleans owes New York $13,000,- 
000. There is a "balance of trade" of $2,750,000 against New 
Orleans and in favor of New York. Hence, the demand in New 
Orleans for drafts on New York is greater than the demand in 
New York for drafts on New Orleans and, therefore, the drafts 
are at a premium in New Orleans. But if New York owes New 
Orleans $13,000,000 and New Orleans owes New York $10,250,- 
000, the "balance of trade," $2,750,000, is against New York and 
in favor of New Orleans. Hence, the demand in New Orleans 
for drafts on New York is less than the demand in New York 
for drafts on New Orleans and, therefore, the drafts are at a dis- 
count in New Orleans. 

If the trade between the two places is the same, the rate of ex- 
change is at par. 

The reason why the banks in New York should charge a pre- 
mium, when the balance of trade is against them, is that they 
must be at the expense of actually sending money to the New 
Orleans banks or be charged interest on their unpaid balance ; 
the reason why the New Orleans banks will sell at a discount is 
that they are willing to sell for less than the face of a draft in 
order to get the money owed them in New York immediately. 

Exchange is charged from -J to %, and is designed to cover 
the cost of transporting the funds from one place to another. 



136 



FTNKEL'S SOLUTION BOOK. 



I. 



II. 



III. 
I. 



!!. 



What will a 30 days' draft on New Orleans for $7216.85 
cost, at % discount, interest 6% ? 

1. 100 %=face of draft. 

2. -f%=discount. 

3. 100% f %=99f %=face less the discount. 

4. 6%=bank discount for 1 year. 

5. %v%~'ffir f 6%=bank discount for 33 days. 

6. 99f % -i^%=99 3 ^%=cost of the draft. 

7. 100%=47216.85, 

8. l%= T fo. of $7216.85=$72.1685, and 

9. 99^ ^^99^. t i mes $72.1685=$7150.094=cost of the 

draft. 

.-. $7150.094=cost of the draft. 

The aggregate face value of two notes is $761.70 and each 
has 1 year 3 months to run; one of the notes I had dis- 
counted at 10% true discount and the other at 10% 
bank discount, and realized from both notes $671.50. 
Find the face value of both notes. 

100%=face of note discounted at bank discount. 
$761.70 100%=face of note discounted at true 

discount. 

10% bank discount for 1 year. 
12^%=bank discount for 1 year 3 months. 

1. 100%=present worth of second note. 

2. 10%=interest on present worth for 1 year. 

3. 12%=interest for 1 year 3 months. 

4. 100%+12|%=112i%=amount of present worth. 

5. $761.70 100%=amount of the present worth. 



(1.) 

(2.) 

(3.) 
(4.) 



(5.) 



(6.) 

(70 

(8.) 

(9.) 

(10.) 

(11.) 
(12.) 

(13.) 



6. .-. 112i%=$761.70 100%, 



of ($761.70 100% )=:$6.7706f 1%, 






100%=100 times ($6.7706| 1%) = $677.06| 

88f %=present worth. 
$761.70 100% ($677.06| 88f % ) = $84.63-J 

H^.% true discount. [discount. 

$84.63i ll^-% t +12i%=$84.63i+ 1 T V % = whole 
$761.70 $671.50=$90.20=whole discount. 



l%=j r of $5.56|=$4.008, and 

100%=100 times $4.008=$400.80=face of note 

discounted at bank discount. 
$761.70 100%==$761.70 $400.80=$360.90=face 

of note discounted at true discount. 



III. 



. ( $400.80=face of note discounted at bank discount, and 
( $360.90=face of note discounted at true discount. 



MISCELLANEOUS PROBLEMS. 



137 



II. 



II. 



3.) 

(*) 

(5.) 
(6.) 
(V.) 

19 

10.) 



1. A merchant bold part of his goods at a profit of 20%, and 
the remainder at a loss of 11%. His goods cost $1000 
and his gain was $100; how mnch was sold at a profit? 
100% cost of goods sold at a profit. Then 
$1000 100% cost of goods sold at a loss. 
20% profit on 100%, the part sold at a profit 

1. 100% $1000 100%. 

2. 1% -j-^of ($1000 100%) $10 1%, 

3. 11% 11 times ($10 1%) $110 11% loss on 

the remainder. 

.'. 20% ($110 11%) 31% $110=gain. 
$100 gain. 
/. 31% $110 $100. 
31% $210, 

1 % ^ T of $210 $6ff , [profit. 

100% 100 times $6ff 677.41ff part sold at a 
III. .-. $677.41f-f value of the part sold at a profit. 

I. By discounting a note at 20% per annum, I get 22-J% 
per annum interest; how long does the note? 

1. 22-% of the proceeds=20% of the face of the note. 

2. 1% of the proceeds f of 20%=|% of the face of the 

note. ** 

3. 100% of the proceeds=100 timss |%=88|% of the face 

of the note. 

4. 100% face of the note. 

5. 88f% proceeds. 

6. 100% 88f %=ll%=discount for a certain time. 

7. 20% discount for 360 days. 

8. 1% discount for -fa of 360 days, or 18 days. 

9 lli%==discount for 11^ times 18 days, or 200 days. 
III. The note was discounted for 200 days. 

I. A man bought a farm for $5000, agreeing to pay princi- 
pal and interest in 5 equal annual installments. What 
will be the annual payment including interest at 6%? 

1. 100 c /c=on& annual payment. 

2. .-. 100%=amount paid at end of the fifth year 

since the debt was then discharged. 

3. 100 %-principal that drew interest the fifth year. 

4. 6 %= interest on this principal. 

5. .*. KXH+6%=106%=amount of this principal. 

6 . . ' . 1 06 #F=100 % - the annual payment . 

7. l %=^ of 100 #= f %, and 

8. 100^=100 times f|%=94f %= principal at the 

beginning of the fifth year. 

9. 94J| %-hlOO $>=194jf ^-amount at the end of the 

fourth year. 



II 



138 



FINKEL'S SOLUTION BOOK. 



II. 



(2.) 



(3.) 



'1. 100%=principal at the beginning of the fourth 
year. 

2. 6%=interest on this principal. 

3. 100%+6%=106%=amount. 

4. .". 106%=194%, 

5. 1%= T ^ of 194ff%==1.83 Y 9 5 -^g-%, and 

6. 100%=100 times 1.83^ r %%=183^\ 3 %=princi- 

pal at the beginning of the fourth year. 

7. lS&f^%+lQO%=285Jffo%==amount at the end 

of the third year. 

1. 100%=principal at the beginning of the third 

year. 

2. 6%=interest. [third year. 

3. 100%+6%=106% = amount at the end of the 

4. .-. 106%=283^ftft F %, 

6. 100%3oO time^F67 T VAVT% ? ^67 T 4 ^ r V% = 

principal at the beginning of third year. 

7. 267AWTV%+100% 367 T 4 A 8 8rV%^ amount at 

the end of second year. 

1. 100%=principal at the beginning of second year. 

2. 6%=interest [year. 

3. 100%+6%=106%=amount at the end of second 

4. .-. 106%=367 T \\ 8 -8 4 rr%> 

6. 100%=100 times 3.46fff|m%=346|fff|-|-f %= 
principal at the beginning of the second year. 

the end of first year. 

1. 100% principal at the beginning of the first 

year, or the cost of farm. 

2. 6%= interest. 

3. 100%+6%=106%=amount at end of first year. 

4. . 

5. 1 

6. 100%=100 times 4.2; 

%=cost of the farm. 
(6.) $5000=cost of the farm. 

(8.) l%=$5000^421^\^% 4 4%V z: =$ll- 8 698-|-, and 
(9.) 100%=100 times $11.8698=$1186.98+=the an- 
nual payment. 

III. .'. $1186.98+ =the annual payment. 

(Milne's Prac., p. 361, prob. 63.) 



(4.) 



(5.) 




I. A and B have $4700 ; f 

~i% 

B's share; how much has each? 



of A's share equals 



of 



PROPORTION. 



II. 




of B's, 



8. 1% of A's= - of !%%=%%% of B's, and 

9. 100% of A's=100 times fo%=74X% of B's. 

10. 100% B's share. 

11. 74^ 2 T %=A's share. 

12. 100%+742 2 T %=1742 2 T %==sum of their shares. 

13. $4700=sum of their shares. 

14. .-. 174 2 2 T % =$4700, 



15". l== 



>f$4700==$27, and 

16. lOOftf^lOO times $27=$2700=B's share. 

17. 74^ 2 T %=74 Y 2 T times $27 $2000=A's share. 
( $2700=B's share and 

' "I $2000=A's share. 



1. J 



CHAPTER XVI. 

RATIO AND PROPORTION. 



1. Rdtio is the relative magnitude of one quantity as com 
pared with another of the same kind; thus, the ratio of 12 apples 
to 4 apples is 3. 

The first quanity, 12 apples, is called the Antecedent, and the 
second quantity, 4 apples, the consequent. Taken together they 
are called Terms of the ratio, or a. Couplet. 



the common sign of 



2. The Sign of ratio is the colon, 
division with the horizontal line omitted. 

Note. Olney says, "There is a common notion among us, that the French 
express a ratio by divding the consequent by the antecedent, while the En- 
glish express it by dividing the antecedent by the consequent. Such is not 
the fact. French, German, and English writers agree in the above defini- 
tion. In fact, the Germans very generally use the sign : instead of -f-; and 



140 FINKEL'S SOLUTION BOOK. 

by all, the two signs are used as exact equivalents." Some writers, 
however, divide the consequent by the antecedent, as a : b This is ac- 
cording to Webster's definition and illustration. To my mind, to divide 
the antecedent by the consequent is more simple and philosophical and 
should be universally adopted by all writers on mathematics. 

3. A Direct JKatio is the quotient of the antecedent di- 
vided by the consequent. 

4. An Indirect Hatio is the quotient of the consequent 
by the antecedent. 

5. A ratio of Greater Inequality is a ratio greater than 
unity; as, 7:3. 

6. A ratio of Less Inequality is a ratio less than unity; 
as, 4 :5. 

7. A Compound Hatio is the product of the correspond- 
ing terms of several simple ratios. Thus, the compound ratio of 
1 : 3, 5 :4, and 7 : 2 is 1x5x7: 3X4X2. 

8. A Duplicate Ratio is the ratio of the squares of two 
numbers. 

9. A Triplicate Hatio is the ratio of the cubes of two 
numbers; as, a 3 : b z . 

1C. A SubduplicateJRatio is the ratio of the square 
roots of two numbers; as, \/^: \/~&. 

11. A Subtriplicate Ratio is the ratio of the cube roots 
of two numbers; as, $/^~: $/. 

PROPORTION. 

% 

12. Proportion is an equality of ratios. The equality is 
indicated by the ordinary sign of equality or by the double colon, : :. 
Thus . a : b=c \d, or a : b : : c : d. 

13. The Extremes of a proportion are the first and fourth 
terms. 

14. The Means are the second and third terms. 

15. A Mean Proportional between two quantities is a 
quantity to which either of the two quantities bears the same ratio 
that the mean does to the other of the two. 

16. A Continued Proportion is a succession of equal 
ratios, in which each consequent is the antecedent of the next 
ratio. 

17. A Compound Proportion is an expression of 
equality between a compound and a simple ratio. 



PROPORTION. 141 

A Conjoined Proportion is a proportion which 
has each antecedent of a compound ratio equal in value to its 
consequent. The first of each pair of equivalent terms is an an- 
tecedent, and the term following, a consequent. This is also 
called the "Chain rule." 

What is the ratio of -J to f ? 
-s-f = i X f = |,the ratio. 

What is the ratio of 10 bu. to If bu.? 
10 bu. -i- If bu. = 10 X T V = 7, the ratio. 

What is the ratio of 25 apples to 75 boxes? 
Ans. No ratio ; for no number of times one will produce 
the other 

In a true proportion, we must always have greater : less :: 
greater : less or less : greater : : less : greater. The test for the 
truth of a proportion is that the product of the means equals the 
product of the extremes. 

I. Ifa5-cent loaf weighs 7oz. when flour is $8 per barrel, 
how much should it weigh when flour is $7.50 per barrel? 

It should evidently weigh more. 

.. less : greater : : less : greater. 

$7.50 : $8.00 : : 7oz : (? = 7 T y>z.) 

I. If a staff 3 feet long, casts a shadow 2 feet, how high is 
the steeple whose shadow at the same time is 75 feet? 

Since the steeple casts a longer shadow than the staff, it is evi- 
dently higher than the staff. 

.-. less : greater : : less : greater. 

2 feet : 75 feet : : 3 feet : ( ?=112| feet.) 

I. What number is that which being divided by one more 
than itself, gives -ij- for a quotient? 



II.. 

III. 
I. 


1. 

2. 

3. 
4. 
5. 

6. 

7. 

M 


Let f=number. T 


hen 
. : : 1 : 7, whence 

lumber, 
ed by 3 more than itself gives $ for 


f+1 ^ 01 ^ + - 
7(*)=1(*+1) or 
Y*=f+l; whence 

i=-rV an d 
|=2 times T L=i=i 

i=the number. 

/"hat number divid 
a quotient? 



142 FINKEL'S SOLUTION BOOK. 

1. Let f=the number. Then 
2 

=|- or, putting this in the form of a proportion, 




1-+3 : : 7 : 9. [the product of the extremes. 

= 1 ^-j-21> the product of the means being equal to 
'=4=21, 

6. =of21=5i, and 

7. |=2 times 5i=10i=the number. 
III. .-. 10=the number. 

I. If 7 lb. of coffee is equal in value to 5 lb. of tea, and 3 lb. 
of tea to 13 lb. of sugar, 39 lb. of sugar to 24 lb. of rice, 
12 lb. of rice to 7 lb. of butter, 8 lb. of butter to 12 lb. 
of cheese ; how many lb. of coffee are equal in value to 
65 lb. of cheese? 

1. 7 lb. of coffee=5 lb. of tea, 

2. 3 lb. of tea=13 lb. of sugar, 

3. 39 lb. of sugar=24 lb. of rice, 

4. 12 lb. of rice=7 lb. of butter, 

'^5. 8 lb. of butter=12 lb. of cheese, and 
6. 65 lb. of cheese= ?=39 lb. of coffee, 

7X3X39X12X8X65 
'' 5X13X24X7X12 
III. ^ /. 65 lb. of cheese=39 lb. of coffee. 

I. I can keep 10 horses or 15 cows on my farm ; how many 

horses can I keep if I have 9 cows? 
15 cows : 9 cows : : 10 horses : ?=6 horses. 
10 horses 6 horses=4 horses. 
.-. I can keep 4 horses with the 9 cows. 

I. If 2 oxen or 3 cows eat one ton of hay in 60 days, how 
long will it last 4 oxen and 5 cows? 

2 oxen : 4 oxen : : 3 cows : ?=6 cows. 

.-. 4 oxen eat as mnch as 6 cows. If a ton of hay last 3 cows 
60 days, it will last 6 cows, which are equal to 4 oxen, and 5 
cows, or 11 cows, not so long. 

.-. 11 cows I 3 cows : : 60 days : ?=17 T 3 T days. 

I. If 24 men, by working 8 hours a day, can, in 18 days, dig 
a ditch 95 rods long, 12 feet wide at the top, 10 feet wide at the 
bottom, and 9 feet deep; how many men, in 24 days of 12 hours a 
day, will be required to dig a ditch 380 rods long, 9 feet wide at 
the top, 5 feet wide at the bottom, and 6 feet deep? 

95 rods : 380 rods 

24 days : 18 days 

12 hours : 8 hours _ nA men . p__ 1 2 



12 feet 

10 feet 

9 feet 



9 feet 

5 feet 

6 feet 



PROPORTION. 143 

380X18X8X9X5X6X24 



95X24X12X12X10X9 



=12 men. 



A Louisville merchant wishes to pay $10000, which he 
owes in Berlin. He can buy a bill of exchange in Louis- 
ville on Berlin at the rate of $.96 for 4 reichmarks ; or 
he is offered a circular bill through London and Paris, 
brokerage |% at each place, at the following rates : 
1=$4.90=25.38 francs, and 5 francs=4 reichmarks. 
What does he gain by direct exchange? 

1. $.238=1 mark. 

2. $10000=10000-:-.238=42016.807 marks. 

3. $.24=1 mark, since this is the rate of exchange. 

4. .-. $10084.033=42016.807 times $.24=42016.807 marks 

=direct exchange. 

5. 42016.807 marks=( ?=$10165.38.) 

6. $4.90=1 1% ofl=.99J. 

7. JE1=.99|. times 25.38 fr. 

8. 5 fr.=4 marks. 

42016.807X4.90X5 ,00 . . 

Circular 



II 



10. $10165.38 $10084.033=$81.35 = gain by direct ex- 
change. 

III. .'. $81.35=gain by direct exchange. 

I. A wheel has 35 cogs; a smaller wheel working in it, 26 cogs; 
in how many revolutions of the larger wheel will the 
smaller one gain 10 revolutions? 

1. 35 cogs 26 cogs=9 cogs=what the smaller wheel gains 

on larger in 1 revolution of larger wheel. 

2. 26 cogs passed through the point of contact=l revolu- 

tion of smaller wheel. 

3. 1 cog passed through the point of contact-^ revolu- 

tion of smaller wheel. 

II. ^ 4. 9 cogs passed through the point of contact=-^- revolu- 
tion of smaller wheel. 
5. .'. In 1 revolution of larger wheel the smaller gains -% 

revolution of smaller wheel. 

/. 2^- revolution gained : 10 revolutions gained .' : 1 
revolution of larger wheel : ?=28f revolutions of 
larger wheel. 

III. .-. The smaller wheel will gain 10 revolutions in 28-f revo- 
lutions of larger wheel. 

By analysis and proportion. 

26 cogs passed through the point of contact==l revolution of 
the smaller wheel. 



144 FINKEL'S SOLUTION BOOK. 

35 cogs passed through the point of contact=l revolution of" 
the larger wheel. But when the larger wheel has made 1 revo- 
lution, 35 cogs of the smaller wheel have passed through the 
point of contact. If 26 cogs having passed through the point of 
contact make 1 revolution of the smaller wheel, how many rev-* 
olutions will 35 cogs make? 

By proportion, 26 cogs : 35 cogs : : 1 rev. : ?=l^ 9 -g-rev. 

.-. The smaller wheel makes 1/g- revolutions while the largei 
wheel makes 1 revolution. .-. The smaller gains 1^- revolutions, 
1 revolutions^- revolution. If the smaller wheel gains -fa 
revoluion in 1 revolution of the larger wheel to gain 10 revolu- 
tions on t|ie larger wheel, the larger wheel must make more rev- 
olutions. /. less : greater : .: less : greater. 

fa rev. ; 10 rev. ; : 1 rev. of larger ; ?=28f rev. of larger. 

I. If the velocity of sound be 1142 fee.t per second, and the 
number of pulsations in a person 70 per minute, what 
is the distance of a cloud, if 20 pulsations are counted 
between the time of seeing the flash and hearing the 
thunder? 

1. 1142 ft.=distance sound travels in 1 second. 

2. 68520 ft=60Xll42 ft.=distance sound travels in 1 min., 

or the time of 70 pulsations. 

3. .'. If it travels 68520 feet while 70 pulsations are count- 

ed, it will travel not so far while 20 pulsations are 
counted. 

4. .'. greater : less : : greater : less. [145 yd. 2-iJ- ft. 

5. 70 pul. : 20pul. : : 68520 ft. : ?=19577| ft.=3 mi. 5 fur. 

III. .-. The cloud is 3 mi. 5 fur. 145 yd. 2| ft. distant. 

(/?., 3d p., p. 289,prob. 45.) 

PROBLEMS. 

1. If 3 horses, in ^ of a month eat f of a tori of hay, how long 
will of a ton last 5 horses? 

2. If a 4-cent loaf weighs 9 oz. when flour is $6 a barrel, how 
much ought a 5-cent loaf weigh when flour is $8 per barrel? 

3. A dog is chasing a hare, which is 46 rods ahead of the dog. 
The dog runs 19 rods while the hare runs 17; how far must the 
dog run before he catches the hare? 

4. If 52 men can dig a trench 355 feet long, 60 feet wide, and 
8 feet deep in 15 days, how long will a trench be that is 45 feet 
wide and 10 feet deep, which 45 men can dig in 25 days? 

5. If -J- of 12 be 3 what will of 40 be ? Ans. 15. 

6. If 3 be \ of 12, what will of 40 be? Ans. 6& 



II. 



PROBLEMS. 145 

If 18 men or 20 women do a work in 9 days, in what time 
can 4 men and 9 women do the same work? Ans. 13^ 7 T days. 



8. If 5 oxen or 7 cows eat 3 T 4 T tons of hay in 87 days, in what 
time will 2 oxen and 3 cows eat the same quantity of hay? 

Ans. 105 days. 

9. Divide $600 between three men, so that the second man 
shall receive one-third more than the first, and the third f more 
than the second. 

10. Two men in Boston hire a carriage for $25, to go to Con- 
cord, N. H., and back, the distance being 72 miles, with the 
privilege of taking in three more persons. Having gone 20 
miles, they took in A ; at Concord they took in B ; and when 
within 30 miles of Boston, they took in C. How much shall 
each pay? Ans. First man, $7.609|f; second, $7.609 W ; A 
$5.873 T V ; B, $2,864^ ; and C, $1.041 T V 

11. Three men purchased 6750 sheep. The number of A's 
sheep is to the number of B's sheep as f is to 3^, and 4 times the 
number of C's sheep is to the number of A's sheep as -J is to -J-. 
Find the number of sheep each had. C A's= 

Ans. ] B's = 
( C's = 

12. If $500 gain $10 in 4 months, what is the rate per cent? 

Ans. 6%. 

13. If 12 men can do as much work as 25 women, and 5 wo- 
men do as much as 6 boys ; how many men would it take to do 
the work of 75 boys? Ans. 30 men. 

14. If 5 experienced compositors in 16 days, 11 hours each, 
can compose 25 sheets of 24 pages in each sheet, 44 lines on a 
page, 8 words in a line, and 5 letters to a word ; how many in- 
experienced compositors in 12 days, 10 hours each, will it take 
to compose a volume (to be printed with the same kind of type), 
consisting of 36 sheets, 16 pages to a sheet, 112 lines to the 
page, 5 words to a line, and 8 letters to a word, provided that 
while composing an inexperienced compositor can do only ^ as 
much as an experienced compositor, and that the latter work is 
only f as hard as the former? Ans. 16. 

15. If A can do f- as much in a day as B, B can do f as much 
as C, and C can do |- as much as D, and D can do -| as much as 
E, and E can do f as much as F; in what time can F do as much 
work as A can do in 28 days? Ans. 8. 

16. A starts on a journey, and travels 27 miles a day; 7 days 
after, B starts, and travels the same, road, 36 miles a day; in how 
many days will B overtake A? Ans. 21 days. 



146 FINKEL'S SOLUTION BOOK. 

17. A wheel has 45 cogs ; a smaller wheel working in it, 36 
cogs ; in how many revolutions of the larger wheel will the 
smaller gain 10 revolutions? Ans. 40. 

18. If the velocity of sound be 1142 feet per second, and the 
number of pulsations in a person 70 per minute, what is the dis- 
tance of a cloud, if 30 pulsations are counted between the time of 
seeing a flash of lightning and hearing the thunder? 

Ans. 5| mi. 108 yd. If ft 

19. If William's services are worth $15f- a month, when he 
labors 9 hours a day, what ought he to receive for if months, 
when he labors 12 hours a day? Ans $91.91^. 

20. If 300 cats kill 300 rats in 300 minutes, how many cats 
will kill 100 rats in 100 minutes? Ans. 300 cats. 



CHAPTER XVII. 

ANALYSIS. 

1. Analysis, in mathematics, is the process of solving 
problems by tracing the relation of the parts. 

I. What will 7 lb. of sugar cost at 5 cents a pound? 
Analysis for primary classes. 

If one ponnd of sugar costs 5 cents, 7 pounds will cost 7 times 
5 cents, which are 35 cents. 

I. If 6 lead pencils cost 30 cents, what will one lead pencil 
cost? 

Analysis: If 6 lead pencils cost 30 cents, one lead pencil will 
cost as many cents as 6 is contained into 30 cents which are 5 
cents. 

I. If 8 oranges cost 48 cents, what will 5 oranges cost? 

Analysis: If 8 oranges cost 48 cents, one orange will cost as 
many cents as 8 is contained into 48 cents which are 6 cents; if 
one orange costs 6 cents 5 oranges will cost 5 times 6 cents, which 
are 30 cents. 

I. If a boy had 7 apples and ate 2 of them, how many had he 
left? 

Analysis: If a boy had 7 apples and ate 2 of them, he had 
left the difference between 7 apples and 2 apples which are 5 
apples. 

I. If John had 12 cents and found 5 cents, how many cents 
did he then have? 

Analysis: If John had 12 cents and found 5 cents, he then, 
had the sum of 12 cents and 5 cents which are 17 cents. 



ANALYSIS. 147 

Note. If teachers in the Primary Departments would see that their 
pupils gave the correct analysis to such problems, their pupils would often 
be better prepared for the higher grades. After they are thoroughly ac- 
quainted with the analysis of such questions they may be taught to write 
out neat, accurate solutions with far less trouble than if allowed to give 
careless analysis to problems in the lower grades. 

I. If 4 balls cost 36 cents, how many balls can be bought for 
81 cents? 

Analysis: If 4 balls cost 36 cents, one ball will cost as many 
cents as 4 is contained into 36 cents which are 9 cents; if one ball 
costs 9 cents for 81 cents there can be bought as many balls as 9 
is contained into 81 which are 9 balls. 

Written solution. 

( 1. 36 cents=cost of 4 balls. 
II. \ 2. 9 cents=36 cents-^4=cost of 1 ball. 
( 3. 81 cents=cost of 81-=-9, or 9 balls. 

III. /. If 4 balls cost 36 cents, for 81 cents there can be bought 
9 balls. 

I. What number divided by -| will give 10 for a quotient? 



II. 



1. |r=the number. 

2- *-H=* X4=f=quotient 



3. 10=quotient. 

4. /, 1=10, 

5. =of 10=2, and 

6. =3 times 2=6=the number. 



III. .''. 6=the number required. 

I. $24 is f of the cost of a barrel of wine; what did it cost? 

II. |=cost of the wine per barrel. 
2. \ of cost=$24, 
3. \ of cost= of $24=$8, 
4. | of cost=5 times $8=$40, 

(II. .% $40=cost of wine. 

t. What number is that from which, if you take \ of itself, 
the remainder will be 16 ? 

1. ^=the number. 

2. If f=4=remainder after taking away ^. 

3. 16=remainder. 
.-.$=16, 

5. f=i of 16=4, and 

6. ^=7 times 4=28=the number 

III. .'. 28=the required number. 



148 FINKEL'S SOLUTION BOOK. 

I. A boat is worth $900; a merchant owns of it, and sells % 
of his share ; what part has he left, and what is it 
worth ? 

1. ^=part the merchant owned. 

2. | of !=^=part he sold. 

i , 3. .-. f-AHri &=M=T\=Part he had left. 

- 1 $900=value of T f , or the whole ship. 
R , *. $75= T V of $900=value of ^ of the ship. 

1 3. $375=5 times $75=value of ^ of the ship, or part 
he had left. 



$375=value of it. 

I. A and B were playing cards. B lost $14, which was y 7 ^ 
times f as much as A then had ; and when they com- 
menced, | of A's money equaled fy of B's. How much 
had each when they began to play? 

( 1. ) | of A's money=4 of B's. 

(2. ) J of A's money=| of f =-% of B's. 



II. 



=B's money when they began to play. Then 



(3.) f of A's money=8 times A=M of B ' s - 

(4.) 

(5.) ^|. = A's money when they began. 

1. T f=A's money after winning $14 from B. 

2. $14=what B lost. 

3. T 7 V times f= T 7 ^^part A's money is of $14. 

5. ^=4 of $14=$2, and [$14 from B. 

6. If 15 times $2=$30=A's money after winning 
(7.) .'. $30 $14=$ 16= A's money at first. 



.) 

.) 



(8.) .-. if=$16, from (5), 

^(10.) H=35 times $l=$35=B's money at first. 

$16=A's money at first, and 
$35=B's money at first. 

(Stod. Int. A., p. lll.prob. 30.) 

I. A drover being asked how many sheep he had, said, if to 
J of my flock you add the number 9^-, the sum will be 
99^-; how many sheep had he? 

'1. 4|=the number of sheep. 

2. i+9-^=-J of the number-|-9^. 

3. 99= of the number+9f 

5. 4=99^9-^=90, and 

6. f=3 times 90=270=number of sheep. 

III. .'. He had 270 sheep. 



ANALYSIS. 



149 



I. 



II. 



III. 



I. 






II. 



III. 



I. 



Heman has 6 books more than Handford, and both have 
26; how many have each? 

1. |=number Handford has. Then 

2. |+6=Heman's number. 

3. f +f +6=|+6=number both have. 

4. 26=number both have. 

6. .-. f+6=26 or 

5. 4^266=20. 

7. f=iof20=6, and 

8. f=2 times 5=10=Handford's number. 

9. f+6=16=Heman's number. 

Handford had 10 books, and 

Heman had 16 books. (Stod. Int. A., p. 116,prob. 2.) 

A man and his wife can drink a keg of wine in 6 days, 
and the man alone in 10 days ; how many days will it 
last the woman? 

1. 6 days=time it takes both to drink it. 

2. ^=part they drink in one day. 

3. 10 days=time it takes the man to drink it. 

4. T 1 ^=part he drinks in one day. [day. 

5. .'. ^ 'fu == 'f^ A == T"5 ==: P ar * * ne woman drinks in one 

6. -j~!=what the woman drinks in -^--^=1.5 days. 






II. 



III. 



.. It will take the woman 15 days. 

(/?. Alg. /.,/. 112,prob. 59.) 

A man was hired for 80 days, on this condition: that for 
every day he worked he should receive 60 cents, and 
for every day he was idle he should forfeit 40 cents. At 
the expiration of the time, he received $40. How many 
days did he work? 

'1. $.60=what he receives a day. 

2. $48=80 X$. 60= what he would have icceived had he 

worked the whole time. 

3. $40=what he received. 

4. ... $48_$40=$8= w hat he lost by his idleness. 

5. $1=$.60, his wages,+$.40, what he had to forfeit,= 

what he lost a day. 

6. /. $8=what he lost in 8-f-l, or 8 days. 

. 80 clays 8 days=72 days, the time he worked. 

/. He worked 72 days. 

A ship-mast 51 feet high, was broken off in a storm, and 
| of the length broken off, equaled f of the length re- 
maining; how much was broken off, and how much re- 
mained? 



150 FINKEL'S SOLUTION BOOK. 

1. f of length broken off=f of length remaining, 

2. -J of length broken off= of f=| of length remaining, 

3. f of length broken off=3 times f=f of length remain- 

ing. 

4. f length remaining. 



II. 



II. 



j 



5. f length broken off. 

6. f-f-f=y=whole length. 

7. 51 feet=whole length. 

8. /. V=51 feet, 

9. 4=TV of 51 feet=3 feet, and 

10. f=8 times 3 feet=24 feet, length remaining. 

11. |=9 times 3 feet=27 feet, length broken off. 

24 feet length remaining, and 
27 feet=le*igth broken off. 

I. A boy being asked his age, said, "4 times my age is 24 
years more than 2 times my age;" how old was he? 

1. !=his age. 

2. 4xf=f=4 times his age. 

3. 2Xf=|=2 times his age. 

4. .-. | = |._|_24 years or 

5. | 4^4^24 years. 

6. i=|- of 24 years=6 years, and 

L 7. f=2 times 6 years=12 years, his age. 

III. /. He is 12 years old. (Stod. Int. A., p. 116,prob. 16.) 



If 10 men or 18 boys can dig 1 acre in 11 days, find the 
number of boys whose assistance will enable 5 men to 
dig 6 acres in 6 days. 

1. 1 A.=what 10 men dig in 11 days. 

2. YV A.=what 1 man digs in 11 days. 

3. YTO A.=YT f TTT A. what 1 man digs in 1 day. 

4. Y2" A.=YTTF A.=5 times YYTF A.=what 5 men dig in 1 

day. [days. 

5. fT ^.=2% A.=6 times ^ A.=what 5 men dig in 6 

6. .'. 6 A. T 3 T A.=5 T 8 i A.=what is to be dug by the boys 

in 6 days. 

7. 1 A.=what 18 boys dig in 11 days. 

8. Y*g- A.=what 1 boy digs in 11 days. 

9. Yi~g" ^ == TT f Tff A.=what 1 boy digs in 1 day. 

10. -g^ A.= T f'j A.=6 times T ^-g- A.=what 1 boy digs in 6 

days. 
-11. 5-fa A.=what 5 T 8 Y-:~g^, or 189, boys dig in 6 days. 

III. .'. It will take 198 boys. 

(R. 3d p., O. E.,p. 318,prob. 66.) 



ANALYSIS. 151 

A man after doing | of a piece of work in 30 days, calls 
an assistant; both together complete it in 6 days. In 
what time could the assistant complete it alone? 

1. -|==part the man does in 30 days. 

2. -ffa^ of f part he does in 1 day. 

3. -f=f |=part he and the assistant do in 6 days. 

4. T 1 5=-g- of -|=part he and the assistant do in 1 day. 

& ' T*5 5Tff=rA rTr=Ti7r = =P art the assistant does in 1 

day. 
6. ff$=part the assistant does in -J-J^-i-^^lf days. 



II. 



III. /. It will take the assistant 21f days. 

(R. 3d /., O. E.,p. 318,prob. 71.) 

Explanation^.- Since the man does f of the work before he called on the 
assistant, there remains f 1=, which he and the assistant do in 6 daj s. 
Hence they do of , or ^ of the work in one day. If the man and his^ 
assistant do ^ of the work in 1 day and the man does -$ of the work in 1 
day, the assistant does the difference between^ and ^ which is r of the 
work in 1 day. Hence it will take $8-7-ro> or21f days, to do the work. 

L A person being asked the time of day, replied that it was- 
past noon, and that f of the time past noon was equal 
to f of the time to midnight. What was the time of 
day? 

1. | of the time past noon=f of the time to midnight. 

2. i of the time past noon=4 of -f= of the time to mid- 

night. [midnight. 

3. |> or the time past noon,=4 times -$=-f of the time to 

4. f=time to midnight. Then 



II. 



5. -|=time past noon. 



6. f-j | ^=time from noon to midnight. 

7. 12 hours=time from noon to midnight. 

8. .*. |=12 hours, 

9. -j=7f of 12 hours=l^ hours, and [past noon. 
10. ^=4 times 1^- hours=5^- hours=5 hr. 20 min., time 



III. .-. It is 20 min. past 5 o'clock, P. M. 

(Mztne's Prac. A., p. S60, prob. 47.) 

Note. From 3, we have the statement that the time past noon is of the 
time to midnight. Hence, if f is the time to midnight, is the time past 
noon or if { is the time to midnight, ^ is the time past noon. 

I. A person being asked the time of day, said that ^ of the 
time past noon equals the time to midnight. What is 
the time of day? 



152 



FINKEL'S SOLUTION BOOK. 



1. -5-=time past noon. Then 

2. ^=time to midnight. 

3. 5 [-7.= 1 T 2 =time from noon to midnight. 
U t <J 4. 12 hours=time from noon to midnight. 

5. .'. J 7 2 =12 hours. 

6. JfTz of 12 hours=l hour, and 

7. -J=7 times 1 hour=7 hours=time*past noon. 

III. .-. It is 7 o'clock P. M. 

I. A man being asked the hour of day, replied that of the 
time past 3 o'clock equaled \ of the time to midnight; 
what was the hour? 

1. of the time past 3 o'clock=-J of the time to midnight. 

2. |, or the time past 3 o'clock,=4 times -J=f of the time 

to midnight. 

3. |=:time to midnight. 

4. |-=time past 3 o'clock. 

H.-I 5. f+f ==time from 3 o'clock to midnight. 

6. 9 hours=time from 3 o'clock to midnight. 

7. .-. |=9 hours. 

8. \-=\ of 9 hours=l^ hours, and 

9. |=4 times 1 hours=6 hours=time past 3 o'clock. 
.10. f+3 hours=9 hours, time past noon. 

III. .-. It is 9, o'clock, P. M. 

(Brooks 1 Int. A., p. 156,prob. 17.) 

I. A person being asked the hour of day, replied, f of the 
time past noon equals |- of time from now to midnight 
+2f hours; what was the time? 

1. f of time past noon=J of time to midnight-)-2f hours. 

2. J of time past noon=-J of (f+2| hours )=^ of time to 

midnight-]- 1 hours. [to midnight-f-4 hours. 

3. f, or time past noon,=3 times (^-[-1-j- hours )=-J of time 

4. f=time to midnight. 

I J 5. +4 hours=time past noon. [night. 

6. f-HH~4 hours= : |-(-4 hours=time from noon to mid- 

7. 12 hours=time from noon to midnight. 

8. .'. f-f-4 hours=12 hours. 

9. ==12 hours 4 hours=8 hours, 

10. |-=i of 8 hours=2 hours, and * 

11. |~|-4 hours=6 hours=time past noon. 

III. .-. It is 6 o'clock, P. M. 

(Stod. Int. A., p. 128, Prob. 29.) 

I. A father gave to each of his sons $5 and had $30 remain- 
ing; had he given them $8 each, it would have taken all 
his money; required the number of sons, 



ANALYSIS. 



153 



'1. $8=amount each received by the second condition. 
2. $5=amount each received by the first condition. 
II. { 3. $3=$8 $5= excess of second condition over first, on 
each son. [10 sons. 

,4. /. $30=excess of second condition over first, on 30-i-3, or 

I. .. There were 10 sons. 

I. If 50 lb. of sea water contain 2 Ib. of salt, how much fresh 
water must be added to the 50 lb. so that 10 lb. of the 
new mixture may contain -J lb. of salt. 

1. -J, lb. of salt=what 10 lb. of the new mixture contains. 

2. f , or 1, lb. of salt=what 3 times 10 lb., or 30 lb., of the 
,, i new mixture contain. [mixture contain. 

X 3. 2 lb. of salt=what 2 times 30 lb., or 60 lb., of the new 
4. .-. 60 lb. 50 lb.=10 lb.=quantity of fresh water that 
must be added. 



III. 



I. 



II. 




.-. 10 lb. of fresh water must be added that 10 lb. of the 
new mixture may contain -j- lb. of salt. 

A farmer had his sheep in three fields, f of the number 
in the first field equals f of the number in the second 
field, and f- of the number in the second field equals 
of the number in the third field. If the entire num- 
ber was 434, how many were in each field? 

of number in first field=f of number in second 
field. [second field. 

-J- of number in first field=-J- of f=| of number in 
-| , or number in first field,=3 times f=f of number 

in second field. 

| of number in second field=| of number in third 
field. [in third field. 

J of number in second field=-J- of f =f of numbet 
f, or number in second field,=3 times f=f of num- 
ber in third field. 
(3.) f=number in third field. Then 
(4.) -|=number in second field, and 
(5.) 4=t of number in second field=number in first 
field in terms of number in third field. 

three fields. 

434=number in the three fields. 

... 2^=434, 

_ T of 434=2, and [field. 

[=64 times 2=128=number of sheep in third 
5=72 times 2=144=number of sheep in second 
field. [field. 

1^=81 times 2=162=number of sheep in first 



154 



FINKEL'S SOLUTION BOOK. 



III. 



II. 



I. 



II. 



C 162=number of sheep in first field, 

< 144=number of sheep in second field, and 

( 128=number of sheep in third field. 

(Milne's Prac. A., p. 362, prob. 68.) 



pupils there are 32 girls ; how 
that there may be 5 boys to 4 



In a certain school of 80 
many boys must leave 
girls? 

1. 80=whole number of pupils. 

2. 32 number of girls. 

3. go 32=48=number of boys. 

4. ^=number of girls. Then, since the number of boys are 

to be to the number of girls as 5 : 4, 

5. J=number of boys. But 

6. |=32. 

7. = of 32=8, and 

8. 4=5 times 8=40=number of boys. 

9. .'. 48 40=8=number that must leave that there may be 

5 boys to 4 girls. 

III. .'. 8 bo^s must leave that there may be 5 boys to 4 girls. 



How far may a person ride in a coach, going at the rate 
of 9 miles per hour, provided he is gone only 10 hours, 
and walks back at the rate of 6 miles per hour? 

1. 9 mi.=distance he can ride in 1 hour. 

2. 1 mi.=distance he can ride in -J- hour. 

3. 6 mi.=distance he can 'walk in 1 hour. 

4. 1 mi.=distance he can 'walk in \ hour. 

hr.-)-^ hr.= T 5 g- hr.=time it takes him to ride 1 mi. 
and walk back. [and walk back. 

10 hours=time it takes him to ride 10-7- T 5 ^, or 36, mi. 



5. 



6. 



III. .-. He can ride 36 miles. 

I. A hound ran 60 rods before he caught the fox, and f of 
the distance the fox ran before he was caught, equaled 
the distance he was ahead when they started. How far 
did the fox run, and how far in advance of the hound 
was he when the chase commenced? 

r l. f=distance the fox ran before he was caught. Then 

2. -|=distance he was ahead. 

3. f +f=|=distance the hound ran to catch the fox. 

4. 60 rods=distance the hound ran to catch the fox. 

5. .-. f=60 rods, 

6. -J=^ of 60 rods=12 rods, and [ahead. 

7. f=2 times 12 rods=24 rods=distance the fox was 

8. f=3 times 12 rods=36 rods=distance the fox ran be- 

fore he was caught. 



ANALYSIS. 



155 



( 24 rods=distance the fox was ahead, and 

( 36 rods=distance he ran before he was caught. 

If J of 12 be 3 , what will of 40 be ? 

1. of 12=4. 

2. i of 40=10. By supposition 

3. 4=3. Then 

4. 1=4 of 3=|, and 

5. 10=10 times |=7|. 

.-. of 40=7 -J, on the supposition that of 12 is 3. 

Eight men hire a coach; by getting 6 more passengers, 
the expenses of each were diminished $1|; what do they 
pay for the coach? 

1. f amount paid for the coach. [been only 8 men. 

2. -=amount 1 man would have had to pay, had there 

3. T 1 ? =amount 1 man paid since there were 8 men-f-6 men, 

or 14 men. 

4. .' i TT=T6 Tnr=irs= what each saved. 

5. $lf=what each saved. 

6. .-. 



7. = of $i=$ 7 > and 



II. 



18. ff=56 times $- J 2 ^ == $32|=amount paid for the coach. 

HI. .% $32f=amount paid for the coach. 

(R. H. A., p. Jt.OS.prob. 46.) 

Second solution. 

'1. $lf=amount saved by each man. [the six meu. 

2. $14=8X$lf=amount saved by the 8 men and paid by 

3. .*. $2=J of $14=amount paid by each of the 14 men. 
,4. .-. $32f 14 times $2=amount they paid for the coach; 

.-. They paid $32f for the coach. 

For every 10 sheep I keep I plow an acre of land, and 
allow one acre of pasture for every 4 sheep; how many 
sheep can I keep on 161 acres? 

1. 1 A.=what I plow for every 10 sheep I keep. 

2. T ^A.=what I plow for each sheep I keep. 

3. 1 A.=what I allow for pasture for every 4 sheep I keep. 

4. ^A.=what I allow for pasture for each sheep I keep. 

5. .'. T 1 ir A.-|-JA.=^ F A.=land required for every sheep. 

6. c-. 161A.=land required for 161-r-^r, or 460 sheep. 

.-. I can keep 460 sheep on 161 acres. 

(./?. Alg. I.) p. 112,prob. 64') 
Complete analysis. 

If for every 10 sheep I plow 1 acre, for 1 sheep I plow -^ of 
&n acre ; and' if for every 4 sheep I pasture 1 acre, for 1 sheep, I 



III. 
I. 



II. 



III. 



156 FINKEL'S SOLUTION BOOK. 

pasture ^ of an acre ; hence 1 sheep requires ^A.-j-^A., or ^.n.., 
and on 161 A. I could keep as many sheep as /^A. is contained in, 
161 A., which are 460 sheep. 

I. A man was engaged for one year at $80 and a suit of 
clothes; he served 7 months, and received for his wages 
the clothes and $35; what was the value of the clothes? 

1. i|=value of the suit of clothes. 

2. ff4-$80=wages for 1 year or 12 months. 

3. ^^iej^^ of (ff+$80)=:wages for 1 month. 

4. ^-|-$46|=7 times ( T Vf$6|-)== wages for 7 months. 
!!.<! 5. ff +$ 3 5= wages for 7 months. 

6. /. -if +$35= T ^+$46|-. 

8. T%=-% of $11^= ::= $2^-, and 

9. ff=12 times $2=$28=value of suit of clothes. 



III. /. The suit of clothes is worth 



A lady has two silver cups, and only one cover. The 
first cup weighs 8 ounces. The first cup and cover 
weighs 3 times as much as the second cup; and the sec- 
ond cup and cover 4 times as much as the first cup. 
What is the weight of the second cup and the cover? 

1. 3 times weight of second cup=weight of cover-)- weight 

of first cup, or 8 oz. [2f oz. 

2. 1 times weight of second cup= of weight of cover-J- 

3. |=weight of cover. Then 

4. -J+2f oz. weight of second cup. [cover. 

5. I+J+2J oz.=-|-2f oz. = weight of second cup and 
!!.<{ 6. 32 oz. 4 times 8 oz.=weight of second cup and cover, 

by the conditions of the problem. 

7. .'. i+2f oz.=32 oz. 

8. |=32 oz. 2f oz.=29 oz. 

9. |=i of 29 oz.=7 oz. 

10. f=3 times 7^- oz.=22 oz.=weight of cover. [cup. 

11. l+2f oz.=7l oz.+2| oz.= 10 oz. = weight of second 

. ( 22 oz.=weight of cover, and 
* " ( 10 oz.=weight of second cup. 



I. A steamboat that can run 15 mi. per hr. with the current 
and 10 mi. per hr. against it, requires 25 hr. to go from 
Cincinnati to Louisville and return ; what is the dis- 
tance between the cities? 




ANALYSIS. 157 

1. 15 mi.=distance the boat can travel down stream in 

1 hour. [hour. 

2. 1 mi.=distance the boat can travel down stream in -^ 

3. 10 mi. distance the boat can travel up stream in 1 hr. 
II. < 4. 1 mi.=distance the boat can travel up stream in -fa hr. 

5. /. y^- hr.-l-y 1 ^ hr.=^r hr.=time required for the boat to 

travel 1 mi. down and return. 

6. .'. 25 br.=time required for the boat to travel 25-r-J-, or 

150, milss down and return. 

III. /. The distance between the two places is 150 miles. 

I. A, B, and C dine on 8 loaves of bread ; A furnishes 5 
loaves ; B, 3 loaves; C pays the others 8d. for his share; 
how must A and B divide the money ? 

1. 8 loaves=what they all eat. 

2. 2f loaves what each eats. 

3. .'.5 loaves 2f loaves=2-j- loaves=what A furnished 

towards C's dinner. 

4. .-. 3 loaves 2f loaves=4 loaf=what B furnished to- 

wards C's dinner. 

5. .'. |=-|=A's share, and 

6. /. ^- -J B's share. 

7. | of 8d.=7d.=what A should receive, and 
.8. ^ of 8d.=ld.=what B should receive. 

Ill \ ^ should receive 7d., and 

' I B should receive Id. (R. H. A., p. 403, prob. 42.) 

I. A and B dig a ditch 100 rods long for $100; how many 
rods does each dig, if they each receive $50, and A digs at $.75 
per rod, and B at $1.25? 

There has been a vast amount of quibbling about this problem; 
but a few moments consideration should suffice to settle all dis- 
pute, and pronounce upon it the sentence of absurdity. 

We have given, the whole amount each received and the 
amount each received per rod. Hence, if we divide the whole 
amount each received by the cost per rod, it must give the num- 
ber of rods he digs. But by doing this we receive 50-.75, or 66f 
rods, what A digs and 50-^-1.25, or 40 rods, what B digs, or 
106f rods which is the length of the ditch, and not 100 rods as 
stated in the problem. The length of the ditch is a function of 
the cost per rod and the whole cost, and when they are given 
the length of the ditch is determined. We might propose a 
problem just as absurd by requiring the circumference of a circle 
whose area is 1 acre, and diameter 20 rods. Since the area 
and circumference are functions of the diameter, when either 



158 FINKEL'S SOLUTION BOOK. 

of these are given, the other is determined and should not be 
limited to an inaccurate statement. 

If, in the original problem, A's price per rod increases at a 
constant ratio so that when the ditch is completed he is receiving 
$1 per rod, and B's price constantly decreases until when the 
ditch is completed he is receiving $1 per rod, then the problem 
is solvable, and the result is 50 rods each. 

I. A is 30 years old, and B is 6 years old ; in how many 
years will A be only 4 times as old as B? 

!=B's age at the required time. Then 
|=A's age at the required time. 

3. | f=J=difference of their ages. 

4. 30 years 6 years=24 years=difFerence of their ages. 
II.<5. .*. f 24 years. 

6. -i=-jr of 24 years=4 years. [time. 

7. 1=2 times 4 years 8 years, B's age at the required 

8. .-. 8 years 6 years 2 years=the number of years hence 

when A will be only 4 times as old as B. 

III. .-. In 2 years A will be only 4 times as old as B. 

I. Jacob is twice as old as his son who is 20 years of age ; 
how long since Jacob was 5 times as old as his son? 

1. 20 years=son's age at present. Then 

2. 40 years=Jacob's age at present. 

3. |-:=son's age at required time. Then 

4. y ) =Jacob's age at required time. 

5. .'. l - f f=difference of their ages. 

II.<! 6. 40 years 20 years=20 years=difference of their ages. 

7. ... f=20 years, 

8. -J=^ of 20 years=2^ years, and [time. 

9. |=2 times 2-J years=5 years, son's age at the required 
10. /. 20 years 5 years 15 years=time since Jacob was 5 

times as old as his son. 
HI. .'. 15 years ago Jacob was 5 times as old as his son. 

Remarks. Observe that the difference between any two persons' ages is 
constant, that is, if the difference between A's and B's ages is 7 years now, 
it will be the same in any number of years from now; for, as a year is add- 
ed to one's age, it is likewise added to the other's age. But the ratio of 
their ages is constantly changing as time goes on. If A is 3 years old and 
B 5 years old, A is now as old as B; but in 1 year, A's age will be 4 years 
and B's 6 years; A is then f as old as B. In 7 years, A will be 10 years old 
and B 12; A will then be {, or f , as old as B, and so on. The ratio of any 
two persons' ages approaches unity as its limit. 

I. A fox is 50 leaps ahead of a hound, and takes 4 leaps in 
the same time that the hound takes 3 ; but 2 of the 
hound's leaps equal 3 of the fox's leaps. How many 
leaps must the hound take to catch the fox? 



ANALYSIS. 159 

1. 2 leaps of hound's=3 leaps of fox's. 

2. 1 leap of hound's=J of 3 leaps=l leaps of the fox's. 

3. 3 leaps of hound's=3 times 1 leaps 4^ leaps of fox's, 
j- 1 4. .-. 4-J- leaps 4 leaps= leap=what the hound gains in 

taking 3 leaps. [ing 6 leaps. 

5. .'. 1 leap=2 times leap=what the hound gains in tak- 

6. .'. 50 leaps=what the hound gains in taking 50x6 

leaps, or 300 leaps. 
III. .-. The hound must take 300 leaps to catch the fox. 
Remark We see that 3 of the hound's leaps equals 4| leaps of the fox's, 
But while the hound takes 3 leaps, the fox takes 4 leaps; hence the hound 
gains 4 4, or |, leap of the fox's. But he has 50 leaps of the fox's to gain, 
and since he gains \ leap of the fox's in 3 leaps, he must take 300 leaps to 
gain 50 leaps. 

I. If 6 sheep are worth 2 cows, and 10 cows are worth 5 
horses; how many sheep can you buy for 3 horses? 

1. Value of 2 cows=value of 6 sheep. 

2. Value of 1 cow=value of 3 sheep. 

3. Value of 10 cows=value of 30 sheep. But 10 cows are 



II. 



worth 5 horses, 



4. /. Value of 5 horses=value of 30 sheep. 

5. Value of 1 horse=value of 6 sheep. 

6. Value of 3 horses=value of 18 sheep. 
III. /. 3 horses are worth 18 sheep. 

I. A teacher agreed to teach a certain time upon these con- 
ditions : if he had 20 scholars he was to receive $25; 
but if he had 30 scholars, he was to receive but $30. 
He had 29 scholars. Required his wages. 

1. $25=his rate of wages for 20 pupils. 

2. $1.25=2^ of $25=his rate of wages for 1 pupil. 

3. $30 his rate of wages for 30 pupils. 

4. $l=^j- of $30 his rate of wages for 1 pupil. 

5. /. $1.25 $1.00=$.25=reduction per pupil by the ad- 



II. 



dition of 10 pupils. 



6. $.025 $.25-r-10=reduction per pupil by the addition of 

1 pupil. 

7. $.225=9 times $.025=reduction per pupil by the addi- 

tion of 9 pupils. 

/. $1.25 $.225=$1.025=his rate of wage per pupil. 
9. $29.725=29 times $1.025=his wages for 29 pupils. 
III. .-. His wages were $29.725. 

(Matt(*i?sArith.,p. 385, f rob. 200.) 

Note. This problem is really indeterminate, because there is no definite 
rate of increase of wages given for each additional scholar. We might say, 
since the wages were increased $5 by the addition of 10 scholars, they would 
be increased $.50 by the addition of one scholar and, consequently, $4.50 by 
the addition of 9 scholars. Hence, his wages should be $25+$4.50, or 
$29.50. By assuming different relations between the increase of wages and 
additional scholars, other results may be obtained. The above solution 
seems to be the most satisfactory. 



160 FINKEL'S SOLUTION BOOK. 

I. A gold and silver watch were bought for $160; the silver 
watch cost only ^ as much as the gold one ; how mucb 
was the cost of each ? 

1. ^-=cost of the gold watch. Then 

2. ^=cost of the silver watch. 

3. 7.+i=8=cost of both. 

4. $160=cost of both. 

5. .'. f=$160, 

6. -J-=-J of $160='$20=cost of the silver watch, and 

7. 7._7 times $20=$140=cost of the gold watch. 

I $20=cost of the silver watch, and 
' ( $140=cost of gold the watch. 

A man has two watches, and a chain worth $20; if he put 
the chain on the first watch it will be worth f as much 
as the second watch, but if he put the chain on the sec- 
ond watch it will be worth 2 times the first watch 
what is the value of each watch? 

1. B.=J f.+$20. 

2. -k s.=J- of (f f.+$20)=J- f.+$10. 

3. f s.= 3 times (i f.+$10)= f +$30. [lem. 

4. I s.= 1 ^ f. $20, by the second condition of the prob- 

II. 



5. ... 11 f._$20=f f.+$30, whence 
y f._ 4 f.=f 



6. V f - f f.=$30+$20, or 

7. f f. $50. 

8. i f.=J. of $50=$10, and 

9. | f.=4 times $10=$40=value of first watch. 

10. f s.=f f.+$30t=f of $40+$30=$90=value of the sec- 
ond watch. 
TTT . \ $40=value of first watch, and 
lli< V I $90=value of second watch. 

( White's Comp. Arith., p. 248, prob. 60.) 

I. At the time of marriage a wife's age was -| of the age of 
her husband, and 10 years after marriage her age was 
Y 7 ^ of the age of her husband ; how old was each at 
the time of marriage ? 

1. |=husband's age at the time of marriage. Then 

2. |=wife's age at the time of marriage. 

3. f+10 years=husband's age 10 years after marriage. 

4. f +10 years wife's age 10 years after marriage. But 

5. T 7 u+7 years y 7 ^ of (f+10 years )=wife's age 10 years 
j i after marriage, by second condition of the problem. 

' yVK years=f+10 years. Whence 
f=10 years 7 years, or 

3 years. [of marriage. 

=10 times 3 years=30 years=husband's age at time 
or T 6 Q,=6 times 3 years=18 years=wife's age at the 
time of marriage. 




ANALYSIS. 161 

( 30 years=husband's age at time of marriage, and 
' ' ' (18 years=wife's age at time of marriage. 

( White's Comp. A., p. 2^1, prob. 35.) 

I. Ten years ago the age of A was f of the age of B, and 
ten years hence the age of A will be f of the age of B ; 
find the age of each. 

1. | B's age 10 years ago. Then 

2. |=A's age 10 years ago. 

3. ^+10 years B's age now, and 

4. f+10 years=A's age now. 

5. |+20 years B's age 10 years hence, and 

6. f+20 years=A's age 10 years hence. [hence. 

7. 5. O f (|_|_20 years)=f+16f years=A's age 10 years 

8. /. f +16f years=+20 years ; whence 

9. 5. 1=20 years 16| years, or 

10. -^=3^ years, and 

11. j-f=12 times 3iJ- years=40 years=B's age 10 years ago. 

12. |= T 9 ^=9 times 3 years=30 years=A's age 10 years 

ago. 

13. ... 11+10 years=50 years=B's age now, and 

14. ^ 2 I 10 years=40 years=A's age now. 



TTT i &Q years=B's age, and 

Ui ' '* (40 years=A's age. 



I. Two men start from two places 495 miles apart, and 
travel toward each other ; one travels 20 miles a day, 
and the other 25 miles a day ; in how many days will 
they meet? 

1. J=number of days. 

2. 20 mi.=distance first travels in 1 day. 

3. f X20 mi. distance first travels in f days. 

4. 25 mi.=distance second travels in 1 day. 

TT )& f X25 mi.=clistance second travels in f days. 

6. .'. f X20 mi.+lX25 mi.=f X(20 mi.+25 mi.)=distance 

both travel. 

7. 495 mi.=di stance both travel. 

8. .-. (20 mi.+25 mi.)x|=(45 mi.)Xf=495 mi. Whence 
f =495-1-45=1 l=number of days. 

III. .-. They will meet in 11 days. 

Second solution. 

rl. 20 miles=distance first travels in a day. 

TT J 2. 25 miles=distance second travels in a day. 

3. .'. 45 miles=distance both travel in a day. [days. 

4. .-. 495 miles=distance both travel in 495-7-45, or 11, 



162 FINKEL'S SOLUTION BOOK. 

III. .-. They will meet in 11 days. 

Third solution the one usually given in the schoolroom. 

20+25=45)495(11 days. 
45 
45 
45 

I. Find a number whose square root is 25 times its cube root. 

1. -|=square root of the number. Then 

2. !xf= tne number, because the square root X the square 

root equals the number. 

3. | the cube root of the number. Then 

4. fXf Xf=the number. But 



II. 



5. |=5X(f)- Hence, squaring both sides, 

6. fxf=25X(|Xf). But 

7. f xf=the number, and 

8. f Xf Xf=the number. 

9- .'. f Xf XS=25X(|Xf ) Dividing by (f Xf ), 
10. f=25. 
[11. ... (f )3 = 25 3 =15625. 

III. .-. The number is 15625. (/?. H. A., p. 367,prob. 14.) 



I. A man bought a horse, saddle and bridle for $150 ; the 
cost of the saddle was of the cost of the horse, and the 
cost of the bridle was -^ the cost of the saddle; what was 
the cost of each? 

1. |"f =cost of the horse. Then 

2. ff=^ of ^|=cost of the saddle, and 

3. Y^=\ of T 2 Y=cost of the bridle. 

4. H=!f+T 2 2+TV=co 
IL<5. $150=cost of all. 



and 

7. J-^ySr of $150=$10=cost of bridle. 

8. H=12 times $10=$120=cost of horse. 
19. T \=2 times $10=$20=cost of saddle. 

( $10=cost of the bridle, 
III. .'. ] $20=cost of the saddle, and 
( $120=cost of the horse. 

( White's Comp. A., p. 241, prol>. S9.) 



ANALYSIS. 



163 



I. 



HI. 



A and B perform T 9 ^ of a piece of work in 2 days, when, 
B leaving, A completes it in -J day; in what time can 
each complete it alone? 

1. T 9 ^=part A and B do in 2 days. 

2. ^=4 f A=P ar t A and B do in 1 day. 

3. -fj rV=TiF P art l e ft after B quits, and which A com- 

pletes in ^ day. 

4. j^=4p=part A can do in 1 day. 

5. .'. -|=part A can do in -!~|=5 days. 

7. .' ^=part B can do in | ; |, or 4, days. 

SA can do the work in 5 days, and 
B can do the work in 4 days. 

( White's Comp. A., p. 280, prob. 193.) 



II. 



I. A and B can do a p.iece of work in 12 days, B and C in 9 
days, and A and C in 6 days; how long will it take 
each alone to do the work? 

1. 12 days time it takes A and B to do the work. 

2. /. T 1 - 2 -=part they do in 1 day. 

3. 9 days=time it takes B and C to do the work. 

4. /. ^ part they do in 1 day. 

5. 6 days=time it takes A and C to do the work. 

6. /. ^=part they do in 1 day. 

7 ... _i__|l|_|_i. ==: ^3 = p art A and B, B and C, and A and C 
do in 1 day twice the work A, B, and C do in 1 day. 

8. .*. T f=| of f=part A, B, and C do in 1 day. 

9. i| ^ y-^-^part A, B, and C do in 1 day part B 

and C do in 1 day=part C does in 1 day. 

10. |f=part C does in f|-:- T V or 10f days. 

11. T | ^= T 5 ^^=part A, B, and C do in 1 day part B and 

C do in 1 day=part A does in 1 day. 

12. T |=part A does in -J|-7-y 5 ^=:14-| days. 

13. II i. ^i.^part A, B, and C do in 1 day part A and 

C do in 1 day=part B does in 1 day. 

14. T f==part B does in ^-^=72 days 

( 14|- days=time it takes A, 
III. .-. < 72 days=time it takes B, and 
( 10^- days=time it takes C. 

( White's Comp. A., p. W^prob. 280.) 

I. The head of a fish is 8 inches long, the tail is as long as 
the head and of the body-flO inches, and the body is 
as long as the head and tail ; what is the length of the 
fish? 



164 FINKEL'S SOLUTION BOOK. 

1. =length of body. 

2. 8 in.=length of head. 

3. | 1. of b.+lO in.+8 in.= 1. of b.+18 in.=length of tail, 

4. | 1. of b.=length of head-f-length of tail. 



II. 



5. .-. f 1. of b.=(4 1. of b.+18 in.)+8 in.=i 1. of b.+26 in 



Whence 

6. I 1. of b. i 1. of b.=J- 1. of b.=26 in. 

7. .'. f 1. of b., or length of body ,=2 times 26 in.=52 in. 

1. of b.+18 in.=26 in.+18 iri.=44 in.=length of tail. 
19. /. 52 in.+44 in.+8 in.s=104 in.=length of the fish. 

III. .-. The length of the fish is 104 inches. 

I. Henry Adams bought a number of pigs for $48 ; and 
losing 3 of them, he sold |- of the remainder, minus 2, 
for cost, receiving $32 less than all cost him; required 
the number purchased. 

1. f=remainder after losing 3. Then 

2. f-j-3=number at first. 

3. f of r. 2=number sold. 

4. $48 $32=$ 16= what was received for f of r. 2. 

5. $8= of $16=what was received for -J- of (fofr. 2), 
j or -| of r. 1. 

6. $24=3 times $8=what was received for 3 times (- of 

r l)=|ofr. 3. 
. $48 $24=$24=what (f of r.+3) (f of r. 3), or <3 

pigs cost. 

4= of $24=what 1 pig cost. 
. $48=what 48-T-4, or 12, pigs cost. 

III. .-. He bought 12 pigs. 

(Brooks' Int. A., f. 164, P rob - &) 

I. A bought some calves for $80; and having lost 10, he sold 
4 more than of the remainder for cost and received 
$32 less than all cost; required the number purchased. 

1. |=remainder after losing 10. Then 

2. f-|-10=number purchased. 

3. f of r.-|-4=number sold. [cost. 

4. $80 $32=$48=cost of f of r.+4, since they sold at 
TT I 5. $24=i of $48=cost of | of (f of r .+4)=J of r.+2. 

^6. $72=3 times $24=cost of 3 times (i of r.+2)=-| of 
r.-f-6. [cost. 

7. .'. $80 $72=$8=what (f of r.+10) (f of r.+6), or 4 

8. $2=^ of $8=what 1 cost. 

9. $80=what 80-r-2, or 40 cost. 

III. .-. He bought 40 calves. 

(Brook's Int. A., p. 164., prob. 10.) 



ANALYSIS. 165 

A lost f of his sheep; now if he finds 5 and sells f of 
what he then has for cost price, he will receive $18; 
but if he loses 5 and sells f of the remainder for cost 
price, he will receive $6; how many sheep had he at 
first? 

1. J= the number of sheep he had at first. 

2. f= the number he lost. 

3. -| f=f > the number he had after losing ^. 

4. f-f-5= the number he had after finding 5. 

5. fof (f+5)=^+3, the number he sold. 

6. | 5= the number, had he lost 5. 

7. | of (| 5)=^ 3, the number he would have sold. 

8. $18=what (-2VI-3) sheep cost. 

9. $6= what (-% 3) sheep cost. 

10. ,-. $12=$18 $6=what (-fg+S) sheep (^3) sheep, 

or 6 sheep cost. 

11. $2= of $12= what 1 sheep cost. 

12. $18= what 18-7-2, or 9 sheep cost. But 

13. $18= what (-2\- (-3) sheep cost. 

14. .-. ^\+3 sheep = 9 sheep, or 

15. ^=6 sheep. 

16. y^= J of 6 sheep=l sheep, and 

17. ff =26 times 1 sheep =25 sheep. 
.-. He had 25 sheep at first. 

(Brooks Int. A., p. 165, prob. 15.) 

A man bought a certain number of cows for $200; had he 
bought 2 more at $2 less each, they would have cost 
him $216; how many did he buy ? 

1. $200=cost of cows. 

2. .$216 cost of oiig'inal number of cows-j-2 more. 

3. $216 $200=$16=cost of 2 cows at $2 less per head. 

4. .- $8=4 of $16=cost of 1 cow at $2. less per head. Then 

5. $S+$2=$10=cost of each cow purchased. 

6. $200=cost of 200-r-lO, or 20 cows. 

. . He bought 20 cows. 

(Brooks Int. A., p. 162, prob. 8.) 

A person being asked the hour of day, said, "the time 
past noon is ^ of the time past midnight;" what was 
the hour? 

1. !=time past midnight. 

2. ^=time past noon. 

3. ... I K^itime from midnight to noon. 

4. 12 hours=time from midnight to noon. 

5. .-. |=12 hours. 

6. -J=^ of 12 hours=6 hours=time past noon. 
III. It was 6 o'clock, P. M. 



166 FINKEL'S SOLUTION BOOK. 

I. Provided the time past 10 o'clock, A. M., equals f of the 

time to midnight; what o'clock is it? 
'1. |=time to midnight. Then 

2. |=time past 10 o'clock. 

3. |-{-f =J=time from 10 o'clock to midnight. 
II.<J 4. 14 hours=time from 10 o'clock to midnight. 

5. .*. ^=14 hours. 

6. =4 of 14 hours=2 hours, and [o'clock P. M. 

7. f=3 times 2 hours=6 hours, time past 10 o'clock=4 
III. ' /. It is 4 o'clock, P. M. 

I. At what time between 3 and 4 o'clock will the hour and 
minute hands of a watch be together? 

1. -f=distance the h. h. moves past 3. Then 

2. 2 ^=12xf distance the m. h. moves past 12. 

3. 2 / -f= 2 ^ 2 =distance the m. h. gains on the h. h. 



II. 



4. 15 min.=distance the m. h. gains on the h. h 



5. ... 22 =15 min 

6. -J=^2 f 15 min. |f min. [past 12. 

7. 2 ^=24 times -J-| min. 16^- min.=distance m. h. moves 
III. * .'. It is 16^i min. past 3 o'clock. 



Remark. In problems of this kind, locate the minute hand at 12 and the 
hour hand at the first of the two numbers between which the conditions of 
the problem are to be satisfied. Thus in the above problem, at 3 o'clock 
the minute hand is at 12 and the hour hand at 3. 

The minute hand moves over 60 minute spaces while the hour hand 
moves over 5 minute spaces. Hence the minute hand moves 12 times 
as fast as the hour hand. Since at 3 o'clock the minute hand is at 12 
and the hour hand at 3, and the minute hand moves 12 times as fast as the 
hour hand, it is evident that the minute hand will- overtake the hour hand 
between 3 and 4. So we let f=distance the hour hand moves past 3 until 
it is overtaken by the minute hand. But since the minute hand moves 12 
times as fast as the hour hand, while the hour move f, the minute hand 
moves 12 times f , or ^ 4 . Now the minute hand has moved from 12 to 3-f|, 
or 15 minutes-)-!, Hence the minute hand has gained 15 minutes on the 
hour hand. It has also gained 2 2 4 $, or % ? . .'. \ 2 =15 minutes. 

In solving any problem of this nature, first locate the hands as previously 
stated, and then ask. yourself how far the minute hand must move to meet 
tke conditions of the problem, if the hour hand should remain stationary. 

I. At what time between 6 and 7 o'clock will the minute 
hand be at right angles with the hour hand? 

1. |-=distance h. h. moves past 6. 

2. V=12 times f=distance m. h. moves past 12. 

3. .*. 2 ^ f= 2 ^ 2 =di stance m. h. gains on h. h. 

4. 15 min. or 45 min.=distance m. h. gains on the h. h. 

5. .. 2 ^ 2 =15 min. or 45 min. 

6. %==% of 15 min. or -^ of 45 min.=!~f min. or 2-^- min. 

7. 2 ^=24 times ^-J min. or 24 times 2-^ min.=16 T 4 T min. 

or 49y T min. 

III. .-. The minute hand will be at right angles with the hour 
hand at 16^ min. or 49^ min. past 6 o'clock. 



ANALYSIS. 



167 



Explanation. Locate the minute hand at 12 and the hour hand at 6. Now 
if the hour hand had remained stationary at 6, the minute hand would have 
to move to 3 or 9, i. e., it would have to gain 15 min. or 45 min. While the 
minute hand is moving to 3 the hour hand is moving from 0. So the min- 
ute hand must move as far past 3 as the hour hand moves past 6. Or while 
the minute hand is moving to 9 the hour hand is moving past 6. So the 
minute hand must move as far past 9 as the hour hand is past 6. .'. The 
minute hand must gain 15 minutes in the first case and 45 minutes in the 
second. 

I. At what time between 2 and 3 o'clock are the hour and 
minute hands opposite? 

1. |=distance hour hand moves past 2. Then 

2. 2 ^=distance the minute hand moves past 12, in the 

same time. [hand. 

3. .'. 2 -% f= 2 or 2 =distance minute hand gained on the hour 
!!.<! 4. 40 min.=distance the minute hand gained on the hour 

hand. 

5. .*. 2 Y 2 =40 min. 

6. J=^ of 40 mim.=l-j\- min., and 

7. 2 ^=24 times 1 T 9 T min.=43i 7 y min. 

III. .. It is 43 T 7 T min. past 2 o'clock when the hands are 
opposite. 

Explanation. Locate the minute hand at 12 and the hour hand at 2. Now 
if the hour hand remained stationary at 2, the minute hand would have to 
move to 8 or over 40 minutes in order to be opposite the hour hand. But 
while the minute hand is moving to 8, the hour hand is moving from 2. So 
the minute hand must move as far past 8 as the hour hand is past 2. Since 
\ is the distance the hour hand moves past 2, f must be the distance the 
minute hand must move past 8. Hence the distance the minute hand 
moves is f+40 min. But \ 4 =distance the minute hand moves. .'. V = l+ 
40 min. or ^=40 min. as shown in step 5. 



I. 



II. 



At what time between 3 and 4 o'clock will the minute 
hand be 5 minutes ahead of the hour hand? 

1. | distance hour hand moves while the m. h. is moving 

to be 5 min. ahead. [moves -| . 

2. 2 ^=12 Xf=distance minute hand moves while the h. h. 

3. /. 2 ^ f= 2 T 2 =distance gained by the minute hand. 

4. 15 min.-}-5 mim.=:20 min.=distance gained by the m. h. 

5. .-. % 2 =20 min. 

6. i=^ of 20 min.=4-f min. 

7. V== 24 times TT min21-jSr min. 



III. /.It is 21 T 9 T min. past 3 o'clock. 

Explanation. Locate the minute hand at 12 and the hour hand at 3. 
Now if the hour hand remained stationary at 3, the minute hand would 
have to move to 4 in order to be 5 min. ahead. But while the minute hand 
is moving to 4 the hour hand is moving from 3. Hence the minute hand 
must move as far past 4 as the hour hand moves past 3. But the hour hand 



168 



FINKEL'S SOLUTION BOOK. 



moves | past 3; hence, the minute hand must move f +5 min. past 4, in all, 
|+20 min. Hence, the minute hand gains (|+20 min.) |=20 min. on the 
hour hand. 

Remark. We always find \*, the distance the minute hand moves, for it 
indicates the time between any two consecutive hours. The hour hand 
indicates the hour. 



I. 



II. 



At what time between 4 and 5 o'clock do the hands of a 
clock make with each other an angle of 45 ? 

1. |=distance the hour hand moves past 4. 

2. 2 ^=distance the minute hand moves past 12. 

3. .'. 2 2 4 f= 2 Y 2 = distance the minute hand gains on the 

hour hand. 



III. 



4. 12 min. or 27-J min.=distance gained by minute hand. 

5. /. 2 Y 2 =12 min. or 27 min. [min. 

6. -|=-j-2 of 12-J min. or % of 27-J min.=J|- min. or 1J 

7. 2 ^=24 times -|^ min. or 24 times 1^ min.=13 1 3 r min. 

or 30 min. 

.*. At 13 T 7 T min. past 4 or 30 min. past 4, the hands make 
an angle of 45 with each other. 

Explanation. Locate the minute hand at 12 and the hour hand at 4. 45 
=i of 360. & of 60 min.=7$ min. Hence, that the hands make an angle of 
45 , the minute hand must be either 7 minutes behind the hour hand or 7| 
min. ahead, Now if the hour hand remained stationary at 4, the minute 
hand would have to move over 12 min. or 2| min. past 2. But 
while the minute hand is moving this distance, the hour hand is moving 
past 4. Hence, the minute hand must move as far past 2 min. past 2 as the 
hour hand moves past 4, z'. e., the minute hand moves f+124 min. Hence, 
it gains (f-f-12 min.) 1=12| min. The reasoning for the second result is 
the same as for the first. 

I. At what time between 4 and 5 o'clock is the minute hand 
as far from 8 as the hour hand is from 3 ? 

1. -|=distance the hour hand moves past 4. 

2. 2 ^=12 times f=distance minute hand moves past 

12 in the same time. 

3. .'. V+f= V=distance both move. 

4. 35 min. distance both move. 

5. /. 2 ^ 6 =35 min. 

6. ^=-5-$ of 35 min.=l^ min. 

7. V=?4 times 1-fc min.=32 T \ min. 

1. f=distance the h. h. moves past 4. 

2. 2 Y 4 =distance minute hand moves past 12. 

3. .'. 2 - 2 f := ? 2 2 =di stance the minute hand gains. 

4. 45 min.=distance the minute hand gains. 

5. .. 2 ^ 2 =45 min. 

6. -J=^- of 45 min.=2^y min. 

7. 2 Y 4 =24 times 2^ min.=49 1 1 r min. 

It is 32 T \ min. or 49y T min. past 4 o'clock. 

(7?. H. A., p. 403,prob. 40.) 



II. 



A. 



B. 



III. 




ANALYSIS. 



16& 



Explanation. This problem requires two different solutions. Locate the 
minute hand at 12 and the hour hand at 4. The hour hand is now 6 min- 
utes from 3. If the hour hand remained stationary, the minute hand would 
have to move to 7 to be 5 minutes from 8. But while the minute hand is 
moving to 7, the hour hand is moving past 4. Hence the minute hand must 
stop as far from 7 as the hour hand moves past 4; . <?., if the hour hand 
moves f past 4 the minute hand must stop f from 7. Then the hour hand 
will be 5 minutes+| from 3 and the minute hand will be f-j-5 minutes from 
8. While the hour hand moved f , the minute hand moved 35 min. $ /. ^ 
=35 min. $, whence ^=35 min. .*. 35 min.=distance they both move. 
The second part has been explained in previous problems. 



I. 



At what time between 5 and 6 o'clock is the minute hand 
midway between 12 and the hour hand? When is the 
hour hand midway between 4 and the minute hand? 



rA. 



II. 



B. 



!=distance the hour hand moves past 5. 
2 ^=distance the minute hand moves in the same 

time. 

f +25 min.=distance from 12 to the hour hand. 
-J- of (f-f-25 min.)=-J-[-12-J- min.=distance minute 

hand moves. 



III. 



5. 
6. 

7. 

8 
9. 

A. 
B. 



mn - 

i=-^g- of 12-j- min.=Jf min. 
V=24 times ff min.=13^ min. 
|-=distance the hour hand moves past 5. 
2 ^=distance the minute hand moves in the same 

time 

f+5 min.=distance the hour hand is from 4. 
Y+10 min. =2 times (f+5 min.)=distance the min- 

ute hand is from 4, since the hour hand is midway 

between it and 4. 
20 min.+(f+10 min.)=f-f-30 min.=distance the 

the minute hand is from 12. 



== __ mm ., or 



of 30 min.=l min. 
= 24 times H min.=36 min. 
It is 13-^g- min. past 5 o'clock. 
It is 36 min. past 5 o'clock 



. 
(R. H. A., p. 403,prob. 41.) 



Explanation. Locate the minute hand at 12 and the hour hand at 5. If 
the hour hand remained stationary, the minute hand would have to move 
over \ of 25 minutes, or 12 minutes. But while it is moving over 12$ 
minutes, the hour hand is moving past 4. Hence, the minute hand will 
have to move 12 minutes+^ of the distance the hour hand moves past 4. 
Hence V=i~l~12| minutes, as shown by step 5 of A. In B, if the hour hand 
remained stationary, the minute hand would have to move over 30 minutes, 
i. e., to 6, that the hour hand may be midway between it and 4. But while 
the minute hand is moving to 6 the hour hand is moving past 4. Hence 
the minute hand must move twice as far past 6 as the hour hand moves past 






170 



FJNKEL'S SOLUTION BOOK. 



4. But |=distance the hour hand moves past 4; hence, =distance the min- 
ute hand moves past 6. Hence, 1+30 minutes=distance the minute hand 
moves. .'. ^* f+30 minutes, as shown by step 6 of B. 



I. 



II.< 



At what time between 3 and 4 o'clock will the minute 
hand be as far from 12 on the left side of the dial plate 
as the hour hand is from 12 on the right side? 

distance the hour hand moves past 3. 
=12 times |=distance the minute hand moves in the 
same time. 

3. 2^4_j_| == 2_6_ : ^i s t an ce they both move. 

4. 45 min.=distance they both move. 

^=45 min. 

-^ of 45 min.=l|-| min. 

=24 times 144 min.=41-A- min. 



5. 
6. 

7. 

III. .-. It is 41^ min. past 3. 

Explanation. Locate the minute hand at 12 and the hour hand at 3. If 
the hour hand remained stationary, the minute hand would have to move to 
9 to be as far from 12 on the left side of the dial plate as the hour hand is 
from 12 on the right. But while the minute hand is moving to 9, the hour 
hand is moving past 3. Hence, the minute hand must stop as far from 9 as 
the hour hand moves past 3. Hence, it is evident, they both move 45 
minutes. 



i. 



II. 



A man looked at his watch and found the time to be be- 
tween 5 and 6 o'clock Within an hour he looked 
again, and found the hands had changed places. What 
was the exact time when he first looked? 

|-=distance m. h. was ahead of h. h., or the dis- 
h. moved, since it changed place 



(8.) 



III. .'. 



tance the h. 

with the m. h. [the two observations. 

2 ^=distance the m. h. moved in the time between 
" V 4 4-f := V = distance they both moved. 
60 mm.=distance they both moved. 
.-. 2^=60 min. 

-i=Jg. of 60 min.=2-j^ min. [ahead of h. h. 

f=2 times 2 T \ min.=4 T % min.=distance m. h. was 
1. |^=distance h. h. was past 5, at time of first obser- 
vation. Then [servation. 
2 ^=distance m. h. was past 12 at time of first ob- 
25 min.-}-f+4 T % min.=f+29 1 ^= distance m. h. 

was past 12 at time of first observation. 
4. 
5. 

6. -j ^2 of 29 T 8 3- rrin.=l^- min. 
-7. 2 2 4 24 times 1-^- min.=32 T % min. 

It was 32 T 4 3 min. past 5 o'clock. 




ANALYSIS. 



171 



Explanation. It is clear that the minute hand was ahead of the hour 
hand at the time of the first observation, or else they could not have ex- 
changed places \vithin an hour. Now, we call the distance from the point 
where the hour hand was located at first lothe point where the minrte hand 
was located first, -|. But in the mean time the hour hand has moved lo the 
position occupied by the minute hand and the minute hand ha moved on 
around the dial to "the position occupied by the hour hand, /. c., the hour 
hand has moved | and the minute 12 times f, or V- Hence, they both 
moved ^ 6 . They both moved 60 minutes since the hand moved on around 
the dial to the position occupied by t'he hour hand and the hour hand mov- 
d to the position occupied by the minute hand. . \ 6 60 min. as shown 
in step (5.) The remaining part of the solution has been explained in pre- 
vious problems. 

At a certain time between 8 and 9 o'clock a boy stepped 
into the schoolroom, and noticed the minute hand be- 
tween 9 and 10. He left, and on returning within an 
hour, he found the hour hand and minute hand had ex- 
changed places. What time was it when he first en- 
tered, and how long was he gone? 

((1.) f=distance m. h. was ahead of the h. h. or dis- 
tance it moved. [J. 

2 ^ 4 =distance m. h. moved while the h. h. moved 
2_4_|_| = 2^6 ^13.^^ both moved. 
(4 ) 60 min =distance both moved. 
^=60 min. 

YV f 60 min.=2 T \ min. [was ahead 

(7.) f=2 times 2 T \ min.=4-% min.=distance m. h. 
r l. |=distance h. h. moved past 8- 

2. 2 1 =distance m. h. moved in same time. 

3. 40 min+|+4 T % m in.=|+44 1 ^ min. = dis- 

tance m. h. moved to be 4 T % min. ahead. 



rA. 



II. 




(2.) 
(3.) 



(5.) 
(6.) 



(8.)^ 4. 
5. 

6. -f 2T of 44 T 8 -g- min.=2-i-g- min. [past 8. 

7. 2 T 4 =24 times 2 T | 7 min. = 48^ min.=time 

1. 2 /> di stance thty both moved. 

2. 60 min. distance they both moved. 
B. 3. .-. 2 T 6 =6C min. 

4. ^=2if of 60 min.=2 T \ min. [was gone. 

5. \ 4 24 times 2 T % min.=55 T \ min.=time he 

( A. It was 48 T 9 6 min past 8 o'clock when he first en- 
< tered school room. 

( B. He was gone 55 T \ min. 

Suppose the hour, minute, and second hands of a clock 
turn upon the same center, and are together at 12 
o'clock; how long before the second hand, hour hand, 
and minute hand respectively, will be midway between 
the other two hands ? 



172 



FINKEL'S SOLUTION BOOK. 



'A. 



II. 



B. 



.C. 



1. |=distance the hour hand moves past 12. Then 

2. ^distance the minute hand moves past 12, and 

3. i_4^4_o = 720 times |=distance 
the second hand moves past 
12. 

4. i_4_4_o_2_4 ^ i_4_i_ 6 = distance 

from the minute hand to the 
second hand. 

5. i^u y == uyj == distance 
from the second hand to the 
hour hand. 

2 ^ |- = 2 ^ 2 = distance from 
the hour hand to the second hand. 
1 - 4 Y 1 - 6 + 1 ^ 1 ~H~ =2 ~T J ^ = distance around the dial. 
60 seconds=distance around the dial as indicated 
by one revolution of the s. h. 



6. 

7. 
8. 

9. 
10. 
11. MP==I440 times 




= f 60 S6C 



sec.=30 T \Vr sec. = time 



when s. h is midway between the h. h. and m. h. 
^=distance the hour hand moves past 12. Then 
^distance the minute hand moves past 12, and 

distance the second 
hand moves past 12. 

4. *- f= 2 Y 2 =distance from h. 
h. to in. h. 

5. 2 Y 2 =distance from s. h. to h. 
h., because the h. h. is mid- 
way between them. [12. 

6. 2 ^ f=distance from s. h. to 

7. 1 - J 2 - 4 -+y ) = 1 - 4 #- = distance 
around the dial. 

8. 60 sec.=distance around the dial. 

9. /. 1 4 6 - =60 sec. 




TIQ.A. 



of ^0 sec.=^ sec. 
11. Uyy^^o times T 3 sec.=59|f sec.=time when 

the h. h. is midway between the s. h. and m. h. 
1 |-=distance h. h. moves past 12. Then 

2. 2 ^=distance m. h. moves past 12, and 

3. 1 -^-=distance s h. moves 
past 12. [h. to s. h. 

4. 2 2 4 f= 2 Y 2 =distance from h. 
5: ^distance from m h. to 

s. h. [from 12 to s h. 

6. |- + 2 2 2 + V = V = distance 

7. -JytP 4_e = i_3_9_4 ^ distance 
around the dial. [dial. 

8. 60 sec =distance around the 




ANALYSIS. 



173 



III. ,\ 



= Sec. 



= sec - 



9. 

10. 

11. 14 2 40 = 1440 times ^ sec.=61f|f sec.=time past 

12 when the m. h. will be midway between the 

h, h. and s. h. 

A. The second hand is midway between h. h. and m. 
h. at 30 T \ 9 Y 7 sec. past 12. [at 59f| sec. past 12. 

B. The hour hand is midway between s. h. and m. h. 

C. The minute hand is midway between h. h. and s. 



h. at 



sec. past 12. 



' From m to .$ = TsTm = * Y V = 1 V 6 - And b 7 the condi- 
tion of the problem, the distance from m to s = the distance from m to h. :. 



Explanation. A. We represent the distance moved by the hour hand 
by |, = the space Th. And since the minute hand moves 12 times as fast 
as the hour hand, it moves ^*. The second hand moves 60 times as fast as 
the minute hand or 720 times as fast as the hour hand. From T to h is f 
and from I to m is ^*. .'. From h to m is Tm 7V/ = *f f = ^. From T to 
s is * Y- 
of t 

from w to ^ = 14 l6 + 1 V 6 - 2 2 - We nave seen, already, that the distance 
from h to m is \ 2 . .'. The whole distance around the dial is 2{ y* 2 + 2 2 2 = 2 *g 4 . 

B. From T to k is f. From T to m is V- ' From // to m=TmTh= 
^* i=V' By the condition of the problem, the distance from h to m=the 
distance from .9 to //. /. sT T/i==^ \=*g. From T around the dial to 
the right of s is 1 ^. .'. The whole distance around the dial= 1 V+^ Q = 

1 -CJ5 

C. From T to h is f . From T to m is ^*. .'. From // to m=Z\=*$-. 
By the condition of the problem, the distance from m to .? the distance 
from h to m= ^. .'. From T to s is 1+^+^=^. From T around the 
dial through T to s is ^ s> . .'. The whole distance around the dial 



I. 



II. 






III. 



A sold to B 9 horses and 7 
same price, 6 horses and 



cows for $300; to C, at the 
13 cows, for the same sum; 



what was the price of each? 

1. Cost of 9 horses-f-cost of 7 cows=$300. Then the 

2. Cost of 36 horses-j-cost of 28 cows=$1200, by taking 

4 times the number of each. 

3. Cost of 6 horses-|-cost of 13 cows=$300. Then the 

4. Cost of 36 horses+cost of 78 cows=$1800, by taking 

6 times the number of each. But 

5. Cost of 36 horses+ccst of 28 cows=$1200. 

6. .'. Cost of 50 cows=$600, by subtracting; and 

7. Cost of 1 cow=-^j- of $600=$12. The 

8. Cost of 7 cows=7 times $12=$84. 

9. .-. Cost of 9 horses-=$300 cost of 7 cows=$300 $84 

=$216. The 
10. Cost of 1 horse=! of $216=$24. 

The cows cost $12 apiece, and 
The horses $24 apiece. 



II. 



174 FINKEL'S SOLUTION BOOK. 

I. A man at his marriage agreed that if at his death he 
should leave onlv a daughter, his wife should have ^ of 
his estate, and if he should leave only a son she should 
have \. He left a son and a daughter. What fractional 
part of the estate should each receive, and what was 
each one's portion, if his estate was worth $6591? 
-1. ^=daughter's share. 

2. ^=wife's share. 

3. |=3 times f=son's share. 

4. J_j_|._|_j_i ? 3 =: i_ w ] lo i e estate. 

5. $6591=whole estate. 

6. /. ^=$6591. [estate. 

7. i= T ^ of $6591=$507=daughter's share,=^g- of whole 
|=3 times $507=$1521=wife's share,= T \ of whole es- 
tate, [tate. 

|=9 times $507=$4563=son's share ,=3% of whole es- 

( $507=Y*g- of whole estate=daughter's share. 
III. .-. ] $1521= T 3 of whole estate=wife's share. 
( $4563= T 9 ^ of whole estate=son's share. 

(Milne's Prac. A., p. 362, prob. 74.) 

Note. For a valuable critique, by Marcus Baker, U. S. Coast Survey, on 
this class of problems, see School Visitor, Vol. IX., p. 186. 

I. There is coal now on the dock, and coal is running on 
also from a shoot at a uniform rate. Six men can clear 
the dock in 1 hour, but 11 men can clear it in 20 min- 
utes ; how long would it take 4 men ? 

1. |=what one man removes in 1 hour. Then 

2. 1 g 8 =6 times |=what 6 men remove in 1 hour. 

3. f=-j of |=what 1 man removes in 20 min., or -J- hour. 

4. =H times |=what 11 men remove in % hour. 

5. /. l f =y= wnat runs on in 1 hr -~ i hr.=| hr. 

Then 
II. J 6. = 1 -^-7-f=what runs on in 1 hour. [commenced. 

7. .*. l f |=|=what was on the dock when the work 

8. |=what 4 men remove in 1 hour. 

9. .'. | J=-J=part of coal removed every hour, that was 

on the dock at first. 
10. j=coal to be removed in |-s-=5 hours. 

(R. H. A., p. 406, prob. 90.) 
III. .-. It will take 4 men, 5 hours to clear the dock. 

Explanation. ^ 2 what 6 men remove in 1 hr. and 2 e 2 =what 11 men re- 
moved in hr. In either case the dock was cleared. /. ^ ^r^g 4 
amount of coal that ran on the dock from the shoot in 1 hr. hr , or hr. 
Hence in 1 hr. there will run on, **---% 1*. Since \ run on in 1 hr. and ^ 
the whole amount of coal removed in 1 hr., ^ 2 |, or f must be the 
amount of coal on the dock when the work began. Since |=the amount 4 
men rerrove in 1 hr. and | the amount that runs on the dock in 1 hr., | 
|, or | is the part of the original quantitv removea each hour. Hence, if \ 
is removed in 1 hour | would be removed in | --\, or hours. 



ANALYSIS. 175 

If 12 oxen eat up 3 acres of pasture in 4 weeks, and 21 
oxen eat up 10 acres of like pasture in 9 weeks ; to find 
how many oxen will eat up 24 acres in 18 weeks. 

1. 10 parts (say)=what one ox eats in a week. Then 

2. 120 parts=12XlO parts=what 12 oxen eat in 1 week, 

3. 480 parts=^4x!20 parts=what 12 oxen eat in 4 weeks. 

4. .'. 480 parts=original grass-|-growth of grass on 3 A. 

in 4 weeks. 

5. 144 parts= of 480 parts=original grass-|-growth of 

grass on 1 A. in 4 weeks. 

6. 210 parts=21XlO parts=what 21 oxen eat in 1 week, 

7. 1890 parts=9 X210 parts=what 21 oxen eat in 9 weeks. 

8. .'. 1890 parts=original grass-j-growth of grass on 10 

A. in 9 weeks. 

9. 189 parts= T L of 1890 parts=original grass-fgrowth 

on 1 A in 9 weeks 

10. .'. 189 parts 144 parts=45 parts=growth on 1 A. in 

9 weeks 4 weeks, or 5 weeks. 

11. 9 parts i of 45 parts=growth on 1 A. in 1 week. 

12. 36 parts=4x9 parts=growth on 1 A. in 4 weeks. 

13. .-. 144 parts 36 parts=108 parts=original quantity of 

grass on 1 A. 

14 2592 parts=24xl08 parts=original quantity on 24 A. 
15. 216 parts=24x9 parts=growth on 24 A. in 1 week. 
16 3888 parts=18x216 parts=growth on 24 A. in 18 

weeks. 

17. /. 2592 parts+3888 parts=6480 parts=quantity of 

grass to be eaten by the required oxen. 

18. 180 parts=18X10 parts=what 1 ox eats in 18 weeks. 

19. .' 6480 parts=what 6480-180, or 36 oxen eat in 18 

weeks. 

III. .-. It will require 36 oxen to eat the grass on 24 A. in 18 
weeks. 



Note. This celebrated problem was, very probably, proposed by Sir 
Isaac Newton and published in his Arithmetica Universalis in 1704. Dr. 
Artemas Martin says, "I have not been able to trace it to any earlier work." 
For a full treatment of this problem see Mathematical Magazine, Vol. 1, 
No. 2. 



I. A man and a boy can mow a certain field in 8 hours, if 
the boy rests 3| hours, it takes them 9 hours. In what 
time can each do it? 



176 FINKEL'S SOLUTION BOOK. 

1. 9J hr. 3f hr.=5f hr.=time they both work together in 

the second case. 

2. 8 hr.=time it takes them to do the work. 

3. .-. -J=part they do in 1 hour. 

4. -^=ff=5f times -J part they do in 5f hours. 

8 

IL<[5. .'. || f|= 7 9 a=part the man did in 3| hours, while the 
boy rested. 

6. /. 3 ff = of -^j=part the man did in 1 hour. 

df 

7. .-. |_jj=part the man can do in ^--7-^ or 13-^- hours. 

8. - ^= 1 - 5 =part the boy does in one hour. 

9. .. |-jj=part the boy can do in f ^-r-^V or 20 hours. 
TTT ( It will take the man 13-J hours, and 

* / The boy 20 hours. (R. H. A., p. 402,prob. 30.) 

I. Six men can do a work in 4 days; after working 2 days, 
how many must join them so as to complete it in 3| 
days? 

1. 4 days=time it takes 6 men. 

2. 26 days=6 times 4^- days=time it takes 1 man. 

3. /. ^g- part 1 man does in 1 day. 

4. Yg=$ times gL=part 6 men do in 1 day. 

5. yV=2 times T 3 ^=part 6 men do in 2 days. [days. 

6. T f T 6 ^=y 7 ^=part to be done in 3f days 2 days, or If 



7. Kj=T3TT :=I P art 1 man does in 1^ days. 



II. 



or 10 men can do in 1-| days. 
9. .*. 10 men 6 men=4 men, the number that must join 

them. 

III. /. They must be joined by 4 more men that they may com- 
plete the work in 3| days. R. H. A., p. 402, prob. 34. 

I. From a ten-gallon keg of wine, one gallon is drawn off 
and the keg filled with water ; if this is repeated 4 
times, what will be the quantity of wine in the keg? 
'1. J^z^part drawn out each time. 

2. T 9 ^=part that was pure wine after the first draught. 

3. y 1 ^ of yV r =Tir7= : P ar t : wine drawn off the second draught. 

4. ^ r | 7y =: T 8 ^= part pure wine left, after the second 

draught. [draught. 

5. yg- of y 8 ^ == T -| i-Q- = part wine drawn off at the third 
y 8 ^ yl^^y 7 ^ 9 ^ part pure wine left after the third 

draught. [draught. 

TTT f T 7 Tnnr == rJMir === P ar ^ : w ^ ne drawn off at the fourth 
TTsV- TWff=TWinr= P art P ure wine left after fourth 

draught. [fourth draught. 

9. /. -j^nnr ^ 10 gal.=6.561 gal.=pure wine left after the 



II. 



draught. 

9292 

5. yV of ( ) == ==part wine drawn off at the third 

draught. 

6. (-^) ^r=(^} =part wine left after the third 



PROBLEMS. 177 

III. .*. There will be 6.561 gal. of pure wine in the keg after 
the fourth draught. 

I. In the above problem, how many draughts are necessary 
to draw off half the wine? 

1. T 1 77 ==part wine drawn off at the first draught. 

2. \ I \=^=part wine left after the first draught. 

Q 

3. T V of T 9 T -^ =part wine drawn off at the sec- 

ond draught. 

9 2 9 2 

4> T 9 iF ^TV(T=(T) =part wine left after the second 



II. 



draught. By induction, 

7. ( T ^) n =rpart wine left after the nth draught. 

8. /. 10( T 9 Tr ) n =number of gal. left after the ^th draught. 

9. 5=number of gal. left after the n\h draught. 
10. .'. 10( T 9 Tr ) n =5, whence 

H- (T 9 7y) n =i- Applying logarithms, 

12. n log. T V=log. f 

13. /. n = log. i-r-log. T V=-30103-:-. T.954243=.301030H- 

.045757=6+. 

III. /. In 7 draughts, half and a little more than half of the 
wine will be drawn off. 

PROBLEMS. 

1. A man bought a horse and a cow for $100, and the cow 
cost f as much as the horse; what was the cost of each? 

Ans. horse, $60; cow, $40. 

2. Stephen has 10 cents more than Marthia, and they to- 
gether have 40 cents; how many have each? 

Ans. Stephen, 25/; Marthia, 15/. 

3. A's fortune added to -J of B's fortune, equals $2000; what 
is the fortune of each, provided A's fortune is to B's as 3 to 4? 

Ans. A's, $1200; B's, $1600. 

4. If 10 oxen eat 4 acres of grass in 6 days, in how many 
days will 30 oxen eat 8 acres? Ans. 4 days. 



178 FINKEL'S SOLUTION BOOK. 

5. If a 5-cent loaf weighs 7 oz. when flour is worth $6 a bar- 
rel, how much ought it weigh when flour is worth $7 per barrel ? 

Ans. 

6. A lady gave 80 cents to some poor children; to each boy 
she gave 2 cents, and to each girl 4 cents; how many were there 
of each, provided there were three times as many boys as girls? 

Ans. 8 girls; 24 boys. 

7. Two men or three boys can plow an acre in ^ of a day ; 
how long will it take 3 men and 2 boys to plow it? 

Ans. J-j da. 

8. A agreed to labor a certain time for $60, on the condition 
that for each day he was idle he should forfeit $2, at the expira- 
tion of the time he received $30; how many days did belabor, 
supposing he received $2 per day for his labor? Ans. 22-J days. 

9. The head of a fish is 4 inches long, the tail is as long as 
the head, plus -J of the body, and the body is as long as the head 
and tail ; what is the length of the fish? Ans. 32 inches. 

10. In a school of 80 pupils there are 30 girls; how many boys 
must leave that there may be 3 boys to 5 girls? Ans. 32. 

11. A steamboat, whose rate of sailing in still water is 12 
miles an hour, descends a river whose current is 4 miles an hour 
and is gone 6 hours; how far did it go? Ans. 32 miles. 

12. A man keeps 72 cows on his farm, and for every 4 cows 
he plows 1 acre, and keeps 1 acre of pasture for every 6 cows ; 
how many acres in his farm.? Ans. 30 acres. 

13. A company of 15 persons engaged a dinner at a hotel, but 
before paying the bill 5 of the company withdrew by which each 
person's bill was augmented $-J; what was the bill? Ans. $15. 

14. A man sold his horse and sleigh for $200, and f of this is 
8 times what his sleigh cost, and the horse cost 10 times as much 
as the sleigh ; required the cost of each. 

Ans. horse, $200; sleigh, $20, 

15. A went to a store and borrowed as much as he had, and 
spent 4 cents; he then went to another store and did the same, 
and then had 4 cents remaining; how much money had he at 
first? Ans. 4 cents. 

16. A lady being asked her age, said that if her age were in- 
creased by its ^, the sum would equal 3 times her age 12 years 
ago; what was her age? Ans. 20. 

17. A lady being asked the hour of day, replied that f of the 
time past noon equaled |- of the time to midnight, minus -J of an 
hour; what was the time? Ans. 6 o'clock, P. M. 



PROBLEMS. 179 

18. What is the hour of day if -j- of the time to noon equals 
the time past midnight? Ans. 9 o'clock, A. M. 

19. A person being asked the time of day, said f of the time 
to midnight equals the time past midnight ; what was the time? 

Ans. 9 o'clock, A. M. 

20. A traveler on a train notices that 4J times the number of 
spaces between the telegraph poles that he passes in a minute is 
the rate of the train in miles per hour. How far are the poles 
apart? Ans. 198 feet. 

21. C's age at A's birth was 5-J times B's age, and now is the 
sum of A's and B's ages, but if A were now 3 years younger and 
B 4 years older, A's age would be f of B's age. Find their ages. 

Ans. A's, 72 years; B's, 88 years; C's, 160 years. 

22. In the above problem change the last and to or, and what 
are their ages? Ans. A's, 36 ; B's, 44, and C's, 80. 

23. I have four casks, A, B, C, and D respectively. Find 
the capacity of each, if f of A fills B, f of B fills C, and fills 
T 9g- of D; but A will fill C and D and 15 quarts remaining. 

Ans. A 35 gal., B 15, C 11^, and D 20. 



24. A man and a boy can do a certain work in 20 days : if 
the boy rests 5^ days it will take them 22^- days; in what time 
can each do it? Ans. The man, 36 da. ; the boy, 45 da. 

25. A can do a job of work in 40 days, B in 60 days; after 
both work 3 days, A leaves ; when must he return that the work 
may occupy but 30 days? Ans. 10 days. 

26. If 8 men or 15 boys plow a field in 15 days of 9-J hr., 
how many boys must assist 16 men to do the work in 5 days of 
10 hr. each? ' Ans. 12 boys. 

27. Bought 10 bu. of potatoes and 20 bu. of apples for $11 ; 
at another time 20 bu. of potatoes and 10 bu. of apples for $13 ; 
what did I pay for each per bu. ? 

Ans. Apples 30/, potatoes 50/. 

28. A farmer sold 17 bu. of barley and 13 bu. of wheat for 
$31.55, getting 35/ a bu. more for wheat than for the barley. 
Find the price of each per bu. 

Ans. Barley 90/, wheat $1.25. 

29. After losing f of my money I earned $12; I then spent f 
of what I had and found I had $36 less than I lost; ho\v much 
money had I at first? Ans. $60. 

30. In a company of 87, the children are -J of the women, and 
the women f of the men; how many are there of each? 

Ans. 54 men, 24 women, and 9 children. 



180 FINKEL'S SOLUTION BOOK. 

31. If 4 horses or 6 cows can be kept 10 days on a ton of hay, 
how long will it last 2 horses and 12 cows? Ans. 4 days, 

32. A, B, and C buy 4 loaves of bread, A paying 5 cents, B 
8 cen^s, and C 11 cents. They eat 3 loaves and sell the fourth 
to D for 24 cents. Divide the 24 cents equitably. 

Ans. A 5 cents, B 8 cents, and C 11 cents. 

33. A and B are at opposite points of a field 135 rods in com- 
pass, and start to go around in the same direction, A at the rate 
of 11 rods in 2 minutes and B 17 rods in 3 minutes. In how 
many rounds will one overtake the other? Ans. B 17 rounds. 

34. If a piece of work can be finished in 45 days by 35 men 
and the men drop off 7 at a time every 15 days, how long will it 
be before the work is completed? Ans. 75 days. 

35 A watch which loses 5 min a day was set right at 12 M., 
July 24th. What will be the true time on the 30th, when the 
hands of that watch point to 12? Ans. 12:30-^ P. M. 

36. A seed is planted. Suppose at the end of 3 years it pro- 
duces'a seed, and on each year thereafter each of which when 3 
years old produce a seed yearly. All the seeds produced : do 
likewise ; how many seeds will be produced in 21 years? 

Ans. 1872. 

37. The circumference of a circle is 390 rods. A, B, and C 
start to go around at the same time. A walks 7 rods per minute, 
B 13 rods per minute in the same direction ; C walks 19 rods per 
minute in the opposite direction. In how many minutes will 
they meet? Ans. 195 min. 

38. If 12 men can empty a cistern into which water is run- 
ning at a uniform rate, in 40 min., and 15 men can empty it in 30 
min., how long will it require 18 men .to empty it? 

Ans. 24 min. 

39. Four men A, B, C, and D, agree to do a piece of work in 
130 days. A gets 42d., B 45d., C 48d., and D 5id., for every 
day they worked, and when they were paid each man has the 
same amount. How many days did each work? [da. 
Ans. A 35ffff da., B 33f \\\ da, C 31^H da -> and D 29 fftl 

40. A fountain has four receiving pipes, A, B, C, and D; A, 
B, and C will fill it in 6 hours; B, C, and D in 8 hours; C, D, 
and A in 10 hours; and D, A, and B in 12 hr.: it also has four dis- 
charging pipes, W, X, Y, and Z ; W, X, and Y will empty it in 
6 hours; X, Y, Z in 5 hours; Y, Z, and W in 4 hours ; and Z, 
W, and X in 3 hours. Suppose the pipes all open, and the 
fountain full, in what time will it be emptied? Ans. 6 T V hours. 




ALLIGATION. 181 



CHAPTER XVIII. 

ALLIGATION. 

1. Alligation is the process employed in the solution of 
problems relating to the compounding of articles of different 
values or qualities. 

/I77 . . . ( 1. Alligation Medial. 

2. All^gat^on \ 2 Alli | ation Alternate. 

I. ALLIGATION MEDIAL. 

1. Alligation Medial is the process of finding the mean, 
or average, rate of a mixture composed of articles of different 
values or qualities, the quantity and rate of each being given. 

I. A grocer mixed 120 lb. of sugar at 5/ a pound, 150 lb. at 
6/., and 130 lb. at 10/.; what is the value of a pound of 

the mixture? 

120 lb. @5/=$6.00, 
150 lb. @6/=$9.00, and 
130 lb. 



II 



1. 
2. 
3. 
4. 400 lb. is worth $28.00. 



III. 



5. .-. 1 lb. is worth $28-r-400=$.07=7 cents. 
.'. One pound of the mixture is worth 7 cents. 

, ( Stod. Comp. A., p. 244, pb. 3. ) 

II. ALLIGATION ALTERNATE. 

1. Alligation Alternate is the process of finding in 
what ratio, one to another, articles of different rates of quality or 
value must be taken to compose a mixture of a given mean, or 
average, rate of quality or value. 

CASE I. 

Given the value of several ingredients, to make a compound of 
a given value. 

I. What relative quantities of tea, worth 25, 27, 30, 32, and 
45 cents per lb. must be taken for a mixture worth 28 
cents per lb. 



Dif. 



Bal. 



SOLUTION. In average, 
the principle is, that the 
gains and loses are equal. ' 
We write the average price 
and the particular values 25, 


'25X 
27/ 
30/ 

32X 

45X 


3X 

I/ 

2/ 
4X 
17/ 


2 lb. 
3 lb. 


17 lb. 

31b. 

i 


4 lb. 
1 lb. 


19 lb. 
4 lb. 
31b. 
1 lb. 
3 lb. 



27, 30, 32, and 45 as in the margin. This is only a convenient 



182 FINKEL'S SOLUTION BOOK. 

arrangement of the operation. Now one pound bought for 25/ 
and sold in a mixture worth 28/ there is a gain of 28/ 25/, or 
3/; one pound bought at 27/ and sold in a mixture worth 28/,, 
there is a gain of 28/ 27/, or I/; one pound bought at 30/ and 
sold in a mixture worth 28/ there is a loss of 30/ 28/, or 2/ ; 
one pound bought at 32/ and sold in a mixture worth 28/, there 
is a loss of 32/ 28/, or 4/; and one pound bought at 45/ and 
sold in a mixture worth 28/ there is a loss of 45/ 28/, or 17/V 
Since the gains and losses are equal, we must take the ingredi- 
ents composing this mixture in such a proportion as to make the 
gains and losses balance. We will first balance the 25/ tea and 
the 30, tea. Since we gain 3/ a pound on the 25/ tea, and lose 
2/ on the 30/ tea, how many pounds of each must we take so 
that the gain and loss on these two kinds may be equal? Evi- 
dently, we should gain 6/ and lose 6/. To find this, we simply 
find the L. C. M. of 3 and 2. Now if we gain 3/ on one pound 
of the 25/ tea, to gain 6/, we must take as many pounds as 3/ 
is contained in 6/, which are 2 Ib. If we lose 2/ on one pound 
of the 30/ tea, to lose 6^, we must take as many pounds as 2/ is 
contained in 6/, which are 3 Ib. Next, balance the 25-cent tea 
and the 45-cent tea. The L. C. M. of 3/ and 17/ is 51/. Now 
if we gain 3/ on one pound of the 25-cent ^ea to gain 51/, we 
must take as many pounds as 3/ is contained in 51/ which are 
17 Ib. If we lose 17/ on one pound of the 45-cent tea, to lose 
51/, we must take as many pounds as 17/ is contained in 51/ 
which are 3 Ib. Next, balance the 27-cent tea and the 32-cent 
tea. The L. C. M. of 1^ and 4/ is 4/. If we gain I/ on one 
pound of the 27-cent tea, to gain 4/, we must take as many 
pounds as I/ is contained in 4/, which are 4 Ib. If we lose 4/ 
on one pound of the 32-cenf tea, if balances the gain on the 27- 
cent tea. Placing the number of pounds to be taken of each, 
kind as shown above, and then adding horizontally, we have 19 
Ib. at 25/, 4 Ib. at 27/, 3 Ib. at 30/, 1 Ib. at 32/, and 3 Ib. at 
45/. It is not necessary to balance them in any particular order. 
All that must be observed, is that all the ingredients be used in 
balancing. 

Note. To prove the problem, use Alligation Medial. 



CASE II. 

To proportionate the parts, one or more of the quantities, but 
not the amount of the combination, being given. 

I. How many bushels of hops, worth respectively 50, 60, 
and 75/ per bushel, with 100 bushels at 40/ per bushel, will 
make a mixture worth 65^" a bushel? 



ALLIGATION. 



183 



Dif. 



40/I25/ 



65/. 



50/ 
60X 
75> 



15/ 

r>p 

W? 



Bal. 



2 bu. 






2 bu. 




2 bu. 




2bu. 






2 bu. 


2 bu. 


5 bu. 


3 bu. 


1 bu 


9 bu. 



-100 bu. 
100 bu. 
100 bu. 
.450 bu. 



Dif. { Bal. 



B. 65/. 



40/ 
50/ 
60/ 


25/ 
15/ 


2bu. 


2 bu. 
3bu. 


10/ 



bu. 



1100 bu. 
2bu. 
2bu. 
254 bu. 



SOLUTION. In this solution, we proceed as in Case I. In A,. 
we obtain the relative amounts to be used of each kind, which is 
2 bu. at 40/, 2 bu. at 50/, 2 bu. at 60/, and 9 bu. at 75/. But 
we are to have 100 bu. of the first kind. Hence, we must multi- 
ply these results by 100-i-2, or 50. Doing this, we obtain 100 
bu. at 40/, 100 bu. at 50/, 100 bu. at 60/, and 450 bu. at 75/. 

Since either or both of the balancing columns, except the first, 
may be multiplied by any number whatever without affecting 
the average, it follows that there are an infinite number of re- 
sults satisfying the conditions of the problem. Since we are to 
have 100 bu. at 40/, the first column can be multiplied by 
only 50. 

In B, we have multiplied the first column by 50 and added in 
the results in the other two columns. This gives us 100 bu. at 
40/, 2 bu. at 50/, 2 bu. at 60/, and 254 bu. at 75/. The second 
and third columns may be multiplied by any number whatever. 
But the first must always must be multiplied by 50, because we 
are to have 100 bu. at 40 cents per bushel. 

(R. H. A., p. 338,prob. 2.) 

I. How much lead, specific gravity 11, with ^ oz. copper, 
sp. gr. 9, can be put on 12 oz. of cork, sp. gr. ^, so that the three 
will just float, that is, have a sp. gr. (1) the same as water? 



r 




3 








3 


x| 




8 




1 


9 




Ti 






I J 



oz.=2 Ib. 1 oz. 



12 oz. 



SOLUTION. The specific gravity of any body is the ratio 
which shows how many times heavier the body is than an equal 



184 FINKEL'S SOLUTION BOOK. 

volume of water. Thus, when we say that the sp^inY gravity 
of lead is 11, we mean that a cubic inch, a cubic foot, a cubic 
yard, or any quantity whatever is 11 times as heavy as an equal 
quantity of water. 

Now if a cubic inch (say) of lead be immersed in water, it 
will displace a cubic inch of water ; and since it weighs 11 times 
as much as a cubic inch of water, it displaces Jy of its own 
weight. Hence, to have equal weights of water and lead we 
must take only Jy as much lead as water. Now since a volume 
of water and y^ as much lead have the same weight, and in 
the proper combination have a volume of 1, since the sp. gr. oi 
the combination is 1, there is a loss of 1 Jy, or T ^, in volume on 
the part of the lead. For the same reason, there is a loss of f in 
volume on the part of the copper, and 3 on the part of the cork. 
Balancing, we see that we must take 3 volumes of lead with T - 
volumes of cork, a unit volume of water being the basis, in order 
that the two substances will just float, /. e., have a specific grav- 
ity (1). In like manner, we must take 3 volumes of copper 
with -| volumes of cork. Now since we must always take 3 vol- 
umes of lead for every T -2- volumes of cork, it is evident that the 
weights of the substances are in the same proportion. Hence, 
we may say, we must take 3 oz. of lead with every y-f- oz. of 
cork, and 3 oz. of copper with every -f oz. of cork. 

But we are to have only -J- oz. of copper. Hence, we must 
multiply the second balancing column by some number that will 
give us -J oz. of copper, /. ., we must multiply 3 by some number 
that will give us -J. The number by which we must multiply is 
-|~j-3 -g-. But multiplying -| by -^, we get ^ 4 T oz. of cork. But 
we are to have altogether 12 oz. of cork. Hence we must yet 
have 12 oz. 2 4 T oz.= 3 ^ T oz. To produce this, we must multi- 
ply T ^ by some number that will give 3 f 2 T oz. This number is 
S^O^H^SSJJ. But we must also multiply 3 by $fj. This will 
.give us 39^ oz.=2 lb. 7-- oz. of lead. Hence, we must use 2 Ib. 7-j- 
oz. of lead, so that the three will just float. 

(jR. H. A., p. 339,prob. 7.) 

I. How many shares of stock, at 40%, must A buy, who has 
bought 120 shares, at 74%, 150 shares, at 68%, and 130 
shares, at 54%, so that he may sell the whole at 60%, 
and gain 20% ? 



> 


'(1.) 100 %=the average cost. 
(2.) 20%=gain. 
(3.) 120% the average selling price. 
(4.) 60%=the average selling price. 
(5.) / 120%=60%. 
(6.) 1^=^ of 60%=^%. 
(7.) 100%=100 times ^%=50%, the average 


cost. 



ALLIGATION. 



185' 



II. 




120 shares @ 74%== 

150 shares @ 68%=10200%. 

130 shares @ 54%= 7020%. 
.-. 400 shares are worth 26100%, and 
1 share is worth 26100%-r-400=65i% , the average. 



3. 50% 



40 % 
65i% 



10 % 



shares. 



10 shares. 



X40=< 



-610 shares. 
.400 shares. 



III. .-. He must take 610 shares. (R. H. A., p. 339, prob. 8.) 



Explanation. Since 60% is the average selling price, and his gain is 
it is evident that his average cost is 60%-M.20, or 50%. In step 3, we find 
that the average cost of the 400 shares is 65^%. Hence, the problem is the 
same as to find how many shares at 40%, must A buy who has 400 shares at 
at an average of 65^% so that his average cost will be 50%. Balancing, 
we find that he must take 15 shares at 40% with 10 shares at 65 \%. But 
he has 400 shares at 65^%. Hence, we must multiply the balancing col- 
umn by 400-MO, or 40. This gives 610 shares at 40%. 



CASE III. 

To proportion the parts, the amount of the whole combination 
being given. 







II. 



III. 



How many barrels of flour, at $8, and $8.50, with 300 
bbl. at $7.50, 800 bbl. at $7.80, and 400 bbl. at $7.65, 
will make 2000 bbl. at $7.85 a bbl. ? 



300 bbl. 
800 bbl. 
400 bbl. 



$7.50 a bbl.=$2250. 
$7.80 a bbl.=$6240. 
$7.65 a bbl.=$3060. 





4. ,'. 1500 *bbl. are worth $11550. 

5. $7.85=the average price per bbl. of 2000 bbl. 

6. /. $15700=2000 X$7.85=the value of 2000 bbl. 

7. .'. $15700 $11550=$4150=th e value of 2000 bbl. 1500 

bbl., or 500 bbl. 
8.30=$4150-f-500=the average value of 1 bbl. 



8. . 

9. $8.30 



$8.00 



$.302 bbl.l 



X( 500^-5 )= 



200 bbl. 
300 bbl. 



$8.50 $.203 bbl.| 
5 bbl. 

1. 200 bbl. at $8.00 per bbl. must be taken with 

2. 300 bbl. at $8.50 per bbl. 

(R. H. A., p. 339,prob. 



I. A dealer in stock can buy 100 animals for $400, at the fol- 
lowing rates: calves, $9; hogs, $2; lambs, $1; how many may 
he take of each kind ? 



186 



FINKEL'S SOLUTION BOOK. 



BaL 





$1 


$3i5 lambs. 




3 


10(17 


24 


31 


38 


45 


62 


59 


$4 


$2 


<feoi 
$z, 


5 hogs. 


68 


60 


52 


44 


36 


28 


20 


12 


4 




$9 


$5:3 calves. 


2 calves. 


29 


30 


31 


32 


33 


34 


35 


36 


37 



8 

Explanation. A lamb bought for $1 and sold for $4 is a gain of $3; a 
hog bought for $2 and sold for $4 is a gain of $2; and a calf bought for $9 
and sold for $4 is a loss of $5. We must make the gains and loses equal. 
The L. C. M. of $3 and $5 is $15. If we gain $3 on one lamb to gain $15 
we must take as many lambs as $3 is contained in $15, which are 5 Iambs. 
If we lose $5 on one calf, to lose $15, we must take as many calves as $5 is 
contained in $15, which are 3 calves. The L. C. M. of $2 and $5 is $10. If 
we gain $2 on one hog, to gain $10, we must take as many hogs as $2 is 
contained in $10, which are 5 hogs. If we lose $5 on one calf, to lose $10, 
we must take as many calves as $5 is contained in $10, which are 2 calves. 
Adding the balancing columns, considering them as abstract numbers, we 
have 8 and 7. 8+7=15. 100-i-15=6f. .' Multiplying each balancing 
column by 6|, will give 33 lambs, 33 hogs, and 33 calves. But this result 
is not compatible with the nature of the problem. Hence we must see if 
we can take a number of 8's and a number of 7's that will make 100. By 
trial, we find that two 8's and twelve 7's will make 100. Hence, multiplying the 
first column by 2 and the second by 12, and adding the columns horizon- 
tally, we have for our result, 10 lambs, 60 hogs, and 30 calves. Again, we 
find, by trying three 8's, four 8's, and so on, that nine 8's taken from 100, 
will leave 28 which is four 7's. Hence, nine 8's and four 7's will make 100. 
Then, multiplying the first column by 9 and the second by 4, and adding 
the columns horizontally, we have for a second result 45 lambs, 20 hogs, 
and 35 calves. Now these are the only answers that can be obtained by 
taking an integral number of 8's and integral number of 7's to make 100. 
But other answers may be obtained by taking 8 a fractional number of 
times, and 7 a fractional number of times to make 100. Suppose, for illus- 
tration, we try to take a number of thirds 8 times. We find that 8 taken 6- 
third times and 7 taken 36 third times will make 100. Multiplying the 
first column by f and the second by 3 ^, and adding the columns horizon- 
tally, we have, for a result, 10 lambs, 60 hogs, and 30 calves the same as 
that obtained by taking 8 twice and 7 twelve times. Again, we find, that 
8 taken 13 third times and 7 taken 28-third times will make 100. Multiply- 
ing and adding as before we find that our results are fractional. Hence, we 
can not take a fraction whose denominator is three. It is clear that we 
must take a fraction whose denominator will reduce to unity, when being 
multiplied by 5. Hence, if we try to take 8 a number of fifths times and 7 
a number of fifths times to make 100, our results will all be integral. -By 
trial, we find that 8 taken 3-fifths times and 7 taken 68-fifths times will 
make 100. Multiplying and adding as before, we have, for our results, 3 
lambs, 68 hogs, and 29 calves. Again, we find that 8 taken 10- fifths times 
and 7 taken 60-fifths times, will make 100. Multiplying and adding as be- 
fore, we have, for results, 10 lambs, 60 hogs, and 30 calves. Again, by trial, 
we find that 8 taken 17-fifths and 7 taken 52-fifths times will make 100. 
Multiplying the first column by y and the second by ^, and adding the col- 
umns horizontally, we have, for results, 17 lambs, 52 hogs, and 31 calves. 
Continuing the process, we find nine admissible answers. These are 
the only answers, satisfying the nature of the problem. 



SYSTEMS OF NOTATION. 



187 



CHAPTER XIX. 

SYSTEMS OF NOTATION. 



1. A System of Notation is a method of expressing 

numbers by means of a series of powers of some fixed number 

called the Radix, or Base of the scale in which the different 
numbers are expressed. 






2. The Jtaclioc of any system is the number of units of one 
order which makes one of the next higher. 



3. 



Names of Systems. 


Radix. 


Names of Systems. 


Radix. 


Unitary 


1 


Nonary 


9 


Binary - . - 


2 


Decimal, or Denary 


10 


Ternary 


3 


Undenary 


11 


Quaternary 


4 


Duodenary, 


12 


Quinary 


5 


Vigesimal, 


20 


Senary 


6 


Trigesimal, 


30 


Septenary 


7 


Sexagesimal, 


60 


Octonary - 


8 


Centesimal, 


100 



4. In writing any number in a uniform scale, as many dis- 
tinct characters, or symbols, are required as there are units in the 
radix of the given system. Thus, in the decimal system, 10 char- 
acters are required; in the ternary, 3; viz., 1, 2, and 0; in the 
senary, 6; viz., 1, 2, 3, 4, 5, and 0; and so on. 

5. Let r be the radix of any system, then any number, N, 
may be expressed in the form, 



the co-efficients ,,, 



2 -f- qr-\- s, in which 
are each less than r. 



To express an integral number in a proposed scale : Divide 
the number by the radix, then the quotient by the radix, and so 
the successive remainders taken in order will be the successive 



on 



digits beginning from units place. 



188 FINKEL'S SOLUTION BOOK. 

I. Express the common number, 75432, in the senary system. 

1. 6)75432 

2. 6)12572+0 



II. 



3. 6)2095+2 

4. 6)349+1 



5. 6)58+1 

6. 6)9+4 
L 7. 1+3 

III. .-. 75432 in the decimal system=1341120 expressed in the 

senary system. 

I. Transform 3256 from a scale whose radix is 7, to a scale 
whose radix is 12. 

1. 12)3256 

2. 12)166+4 

3. 12)11+1 



II 



4. 0+8 

III. .-. 3256 in the septenary system=814 in the duodenary 
system. 

Explanation. In the senary system, 7 units of one order make one of the 
next higher. Hence, 3 units of the fourth order = 7X3, or 21, units of the 
third order. 21 units +2 units =23 units. 23-7-12 = 1, with a remainder 
11. 11 units of the third order =77 units of the second order. 77 units 
+5 units=82 unrts. 82+12=6, with a remainder 10. 10 units of the 
second order=70 units of the first order. 70 units +6 units=76 units. 76 
+12= 6, with a remainder 4. Hence, the first quotient is 166, with a re- 
mainder 4. Treat this quotient in like manner, and so on, until a quo- 
tient is obtained, that is less than 12. 

I. What is the sum of 45324502 and 25405534, in the senary 
system ? 

45324502 

25405534 

115134440 

Explanation. 4+2=6. 6-^-6=1, with no remainder. Write the and 
carry the 1. 3+1=4. Write the 4. 5+5=10. 10+6=1, with a remainder 
4. Write the 4 and carry the 1. 5+4+1=10. 10+6=1, with a remain- 
der 4. Write the 4 and carry the 1. 0+2+1=3. Write the 3. 4+3=7. 
7+6+1 with a remainder 1. Write 1 and carry 1 5+5+1=11. ll-r-6= 
1, with a remainder 5. Write the 5 and carry the 1. 2+4+1=7. 7+6= 
1, with a remainder 1. Write 1 and carry 1. The result is 115134440. 

I. What is the difference between 24502 and 5534 in the 
octonary system? 

24502 

5534 

16746 

Explanation. 4 cannot be taken from 2. Hence, borrow one unit from 
a higher denomination. Then (2+8) 6=4. (81) 3=4. 5 from (4+8) 
7. 5 from (3+8)=6. Hence, the result is 16746. 




SYSTEMS OF NOTATION. 189 

I. Transform 3413 from the scale of 6 to the scale of 7. 

1. 7)3413 
II.<[2. 7)310+3 
3. 7)24+3 
2+2 

III. .-. 3413 in the senary system=2233 in the septenary sys- 
tem. 

I. Multiply 24305 by 34120 in the senary system. 

24305 

34120 
530140 
24305 
150032 
121323 
1411103040 

Explanation, Multiplying 5 by 2 gives 10. HK-6 1, with a remainder 
4. Write 4 and carry 1 to the next order. 2 times 00. 0+11. Write 
the 1. 2 times 3=6. 6-4-6=1, with a remainder 0. Write the and 
carry the 1 to the next higher order. 2 times 4=8. 84-19. 9-*-6=l, 
with a remainder 3. Write 3 and carry the 1 to the next higher order. 
2 times 2=4. 4+1=5. Write 5. Multiply in like manner bv 1, 4, and 3. 
Add the partial products, remembering that 6 units of one order, in the sen- 
ary system, uniformly make one of the next higher. 

I. Multiply 2483 by 589 in the undenary system, or system 
whose radix is 11. 

We must represent 10 by some character. Let it be t. 

2483 
589 



1*985 
1/502 
11184 
13322/5 

Explanation. In the undenary system, 11 units of one order uniformly 
make one of the next higher order. 9 times 3=27. 27 -f- 11=2, with a re- 
mainder 5. Write 5 and carry the 2 to the next higher order, or second or- 
der. 9 times 8=72. 72+2=74. 74^-11=6, with a remainder 8. Write 8 
and carry the 6 to the next higher order, or third order. 9 times 4=36. 
36+6=42. 42-5-11=3, with a remainder 9. Write 9 and carry the 3 to the 
next higher order, or the fourth order. 9 times 2=18. 18+3=21. 21-j-ll 
=1, with a remainder /. Write t and carry the 1 to the next higher order. 
Multiply in like manner by 8 and 5. Add the partial products, remember- 
ing that 11 units of one order equals one of the next higher. Wherever 
10 occurs, it must be represented by a single character t. 



190 FINKEL'S SOLUTION BOOK. 

I. Divide 1184323 by 589 in the duodenary system. 

In the duodenary system, we must have 12 characters ; viz., 1, 
2, 3, 4, 5, 6, 7, 8, 9, t, e, and 0. / represents 10 and e, 11. 
589)1184323(2486 



_ 
22/3 



3*32 
39/0 

1523 

1523 

Explanation. In the duodenary system, 12 units of one order make one 
of the next higher. 1184 will 'contain 589, 2 times. Then multiply the 
divisor, 589, by 2 thus: 2 times 9=18. 18-4-12=1, with a remainder 6. 
Write the 6 and carry the 1. 2 times 8=16. 16+1=17. ' 17-*- 12=1, with a 
remainder 5. Write the 5 and carry the 1. 2 times 5=/. t-}-l=e. 
Write the e. Then subtract. 6 from (12+4)=^, 5 from 7=2, and e from (12 



Hence, the first partial dividend is 22 /. Bring down 3. Then 22t3 will 
contain 589, 4 times. Multiply as before. By continuing the operation we 
obtain 2483 for a quotient. 

I. Divide 95088918 by #4, in the duodenary system. 




I. Extract the square root of 11122441 in the senary system. 



11122441(2405 
2X2=_4 
44 ] 312 
304 



hi 

Li 



2X24=52 
2X240=520 5 



42441 
42441 



Explanation. The greatest square in 11 expressed in the senary system 
is 4. Subtracting and bringing down the next period, we have 312 for the 
next partial dividend. Doubling the root already found and finding how 
many times it is contained in 312 expressed in the senary system, we find it 
is 4. Continuing the process the same as in the decimal system, the re- 
sult is 2405. 



SYSTEMS OF NOTATION. 



191 



I. Extract the square root of 11000000100001 in the binary 
system. 



I. 



II. 



III. 



I. 



III. 



11000000100001(1101111 

1 



101 H 


300 
101 




11001 
110101 
1101101 
11011101 


110000 
11001 


1011110 
110101 


10100100 
1101101 


11011101 
11011101 



( Todhunter's Algebra, p. 855, Ex. 28.) 
Find in what scale, or system, 95 is denoted by 137. 

1. Let r=the radix of the system. Then 
2. 



3. r 2 +3r=95 7=88, and 

4. r 2 4-3r4-f=88+f= 3 1, by completing the square. 

5. 7"-f-f=V> by extracting the square root, and 

6. r=y |=y=8, the radix of the system. 

/. 95 is denoted by 137 in the octonary system. 

( Todhunter's Alg., p. 255, prob. 

Find in what system 1331 is denoted by 1000. 

1. Let r=the radix of the system. Then 

2. r 

3. r 4 =1331. Whence 

4. r= 



=11, the radix of the system. 

/. 1331 is denoted by 1000 in the undenary system. 

( Todhunter's Alg., p. 255, prob. 28.) 



192 



FINKEL'S SOLUTION BOOK. 



CHAPTER XX. 

MENSURATION. 

1, Mensuration is that branch of applied mathematics 
which treats of geometrical magnitudes. 

2. Geometrical Magnitudes are lines, surfaces, arid 
solids. 

3. Geometrical Magnitudes. 







D. Solid. 




c. 


Surface. 


w 


A. Lin 


&. 


Oi 4^ 

ri 
P 


03 




3 
ft 


2. Cylinder. 


,1 

3 




i 

, * 

4*. 00 10 


<t 

r- 4^ CO t-0 HJ 


3 

9 

^d c/5 o 


tTp 

003 
C ? 

< S 

^a'a- 
J~" ,^- 




/^A^S 

to i- 1 


to i- 


-i Oi 4^ 03 to 




^ SPOO 


co RpwO^d 


B : | 


5 




OK 


0? 


0* 




P ^ o ' 


"g. P S^"^ 




^> 






N 

I p-'cS 



?asj?p??^a 

' p ^ ' X 3 P 

as^ciS S"CL3 

sss- |s< 
tUB 



H-> 00 bO H- 1 tO H* 



3 t 
arq P 

c r 



o 
p 

I 

o 

I 



cr 



5d ^ 

3" g- 

I 1 

i * CTQ 

I " 

51 = 




MENSURATION. 193 

A Line is a geometrical magnitude having length, with- 
out bread! h or thickness. 

5. A Straight Line is a line which pierces space evenly, 
so that a piece of space from along one side of it will fit any side 
of any other portion. 

6. A Curved Line is a line no part of which is straight. 

7. A Surface is the common boundary of two parts of a 
solid, or of a solid and the remainder of space. 

8. A Plane Surface, or Plane, is a surface which di- 
vides space evenly, so that a piece of space from along one side 
of it will fit either side of any other portion of it. 

9. A Ctirved Surface is a surface no part of which is 
plane. 

10. A Polygon (ttMyatvos, from Ilokuq^ many, and 
angle) is a portion of a plane bounded by straight lines. 



11. A Circle (xtpxos, circle, ring) is a portion of a plane 
bounded by a curved line every point of which is equally distant 
from a point within called the center, 



1. An Ellipse (e'>Ue0) is a portion of a plane bounded 
by a curved line any point from which, if two straight lines are 
drawn to two points within, called the^/bcz, the sum of the two 
lines will be constant. 

13. A Triangle (Lat. Triangulum, from tries, tria, 
three, and angulus, corner, angle) is a polygon bounded by three 
straight lines. 

14. An Angle is the opening between two lines which 
meet in a point. 

II. Straight Angle. 
8. Oblique j I Acute^ 

16. A Straight Angle has its sides in the same line, and 
on different sides of the point of meeting, or vertex. 

17. A Might Angle is half of a Strait Angle, and is 
formed by one straight line meeting another so as to make the 
adjacent angles equal. 

18. An Oblique Angle is formed by one line meeting 
another so as to make the adjacent angles unequal. 

19. An, Acute Angle is an angle less than a right angle. 

20. An Obtuse Angle is an angle greater than a right 
angle. 



194 FINKEL'S SOLUTION BOOK. 

21* A Hight Triangle is a triangle, one of whose angles 
is a right angle. 

22. An Oblique- Angled Triangle is one whose angles 
are all oblique. 

23. An Isosceles Triangle is one which has two of its 

sides equal. 

24. A Scalene Triangle is one which has no two of its 

sides equal. 

25. An Equilateral Triangle is one which has all 
the sides equal. 

26. A Quadrilateral (Lat. quadrilaterus , from quatuor^ 
four, and latus, lateris, a side) is a polygon bounded by four 
straight lines. 

21. A Parallelogram ( flap ally 16? pawov, from 
IlapdMyJios, parallel, and rpafi.fjt.rj, a stroke in writing, a line) 
is a quadrilateral having its opposite sides parallel, two and two. 

28. A Hight Parallelogram is a parallelogram whose 
angles are all right angles. 

29. An Oblique Parallelogram is a parallelogram 
whose angles are oblique. 

30. A Rectangle (Lat. rectus, right, and angulus, an 
angle) is a right parallelogram. 

31. A Square is an equilateral rectangle. 



32. A ^Rhomboid (fafjLfioetd&s, from ^o^/foc, rhomb, and! 
shape) is a parallelogram whose angles are oblique. 



33. A Rhombus fitppos, from jMpfieiv, to turn or whirl 
round) is an equilateral rhomboid. 

34. A Pentagon ( Ilsvrd^wvov, /7^re, five, and futvia > 
angle) is a polygon bounded by five sides. Polygons are named 
in reference to the number of sides that bound them. A Hexa- 
gon has six sides; Heptagon, seven; Octagon, eight; Nonagon r 
nine; Decagon, ten; Undecagon, eleven; Dodecagon, twelve; 
Tridecagon, thirteen ; Tetradecagon, fourteen; Pentedecagon, 
fifteen; Hexdecagon, sixteen; Heptadccagon, seventeen; Octa- 
decagon, 'eighteen; Enneadecagon , nineteen; Icosagon, twenty; 
Icosaisagon, twenty-one ; Icosad&agon, twenty -two; Icosatriagon, 
twenty-three \Icosatetragon, twenty-four ; Icosapentegon , twenty- 
five; Icosakexagon, twenty-six; Icosaheptagon, twenty-seven; 
Icosaoctagon, twenty-eight ; Icosaenneagon, twenty-nine; Tria- 
contagon, thirty; Tficontaisagon, thirty-one; Tricontadoagon, 
thirty-two; Tricontatriagon, thirty-three; and so on to Tessa- 
racontagon, forty; Pentecontagon, fifty; Hexacontagon, sixty;: 






MENSURATION. 195 



Hebdomacontagon , seventy ; Ogdoacontagon, eighty ; Enenacon- 
tagon, ninety; Hecatonagon , one hundred; Diacosiagon, two 
hundred; Triacosiagon, three hundred; Tetracosiagon, lour hun- 
dred; Pentecosiagon, five hundred; Hexacosiagon, six hundred; 
Heptacosiagon, seven hundred ; Oktacosiagon, eight hundred ; 
Enacosiagon, nine hundred; Chiliagon, one thousand; &c. 

35. A Spherical Surface is the boundary between a 

sphere and outer space. 

36. A Conical Surface is the boundary between a cone 
and outer space. 

37. A Cylindrical Surface is the boundary between 
the cylinder and outer space. 

38. A Solid is a part of space occupied by a physical body, 
or marked out in any other way. 

39. A Polyhedron (IJoMedpoq, from 77oAy^, many, and 
seat, base) is a solid bounded by polygons. 

40. A Prism is a polyhedron in which two of the faces 
are polygons equal in all respects and having their homologous 
sides parallel. 

41. The Altitude of a prism is the perpendicular distance 
between the planes of its bases. 

4:2. A Triangular Prism is one whose bases are trian- 
gles. 

43. A Quadrangular Prism is one whose bases are 
quadrilaterals. 

44. A Parallelopipedon is a prism whose* bases are 
parallelograms. 

45. A Hight Parallelopipedon is one whose lateral 
edges are perpendicular to the planes of the bases. 

46. A Rectangular Parallelopipedon is one whose 
faces are all rectangles. 

47. A Cube x6/5o<?, a cube, a cubical die) is a rectangular 
parallelopipedon whose faces are squares. 

48. A Hight Prism is one whose lateral edges are per- 
pendicular to the planes of the bases. 

49. An Oblique Prism is one whose lateral edges are 
oblique to the planes of the bases. 

50. A Pyramid (flupa^) is a polyhedron bounded by a 
polygon called the base, and by triangles meeting at a common 
point called the vertex of the pyramid. 



196 FINKEL'S SOLUTION BOOK. 

51. The Convex Surface of a pyramid is the sum of the 

triangles which bound it. 

58. A Hight Pyramid is one whose base is a regular 
polygon, and hi which the perpendicular, drawn from the vertex 
to the plane of the base, passes through the center of the base. 
The perpendicular is called the axis. 

53. A Tetrahedron (rirpa four, and Ifl/oa, seat, base) 
is a pyramid whose faces are all equilateral triangles. 

54. The Altitude ( Lat. Altitudo, from altus, high, and 
ude denoting state or condition) of a pyramid is the perpendicu- 
lar distance from the vertex to the plane of the base. 

55. The Slant If eight of a pyramid, is the perpendicu- 
lar distance from the vertex to any side of the base. 

56. A Triangular Pyramid is one whose base is a 
triangle. 

57. An Octahedron (dxrdsdpos from dxrtf eight, and 
tSpa seat, base) is a polyhedron bounded by eight equal equi- 
lateral triangles. 

58. A Dodecahedron (dudsza, twelve, and gdpa, seat, 
base) is a polyhedron bounded by twelve equal and regular 
pentagons. 

59. An Icosahedron (efxo?*, twenty, and Zdpa, seat, base) 
is a polyhedron bounded by twenty equal equilateral triangles. 

60. A Cylinder (xuhvdpoz, from xuMvdew, xvUew, to 
roll) is a solid bounded by a surface generated by a line so mov- 
ing that every two of its positions are parallel, and two parallel 
planes. 

61. The Axis (a^utv) of a cylinder is the line joining the 
centers of its bases. 

62. A Hight Cylinder is one whose axis is perpendicu- 
lar to the planes of the bases. 

63. A Cone (xwvos, from Skr. co, to bring to a point) is 
a solid bounded by a surface generated by a straight line moving 
so as always to pass through a fixed point called the apex,, and a 
plane. 

64. A Hi(/ht Cone is a solid generated by revolving a 
right-angled triangle about one perpendicular. 

65. An Oblique Cone is one in which the line, called 
the axis, drawn from the apex to the center of the base is not 
perpendicular. 

66. The Frustum (Lat. frustum, piece, bit) of a pyra- 
mid or a cone is the portion included between the base and a 
parallel section. 

67. A Sphere (ff^aipa) is a solid bounded by a curved 




MENSURATION. 



197 






surface, every point of which is equally distant from a point 
within, called the center. 

Before we enter into the solution of problems in Mensuration, 
it will be necessary first to explain a difficulty which we en- 
counter. 

The common way of teaching that feet multiplied by feet give 
square feet is wrong ; for there is no rule in mathematics justify- 
ing the multiplication of one denominate number by another. 
If it is correct to say feet multiplied by feet give square feet, we 
might, with equal propriety, say dollars multiplied by dollars 
give square dollars a product wholly unintelligible. In all our 
reasoning, we deal with abstract numbers alone or the symbols 
of abstract numbers. These do not represent lines, surfaces, or 
solids, but the relations between these numbers may represent 
the relations between the magnitudes under consideration. 

Suppose, for example, that the line AB contains 5 units, and 
the the line BC 4 units. Let a denote the abstract number 5, 
and b the abstract number 4. Then 
#3=20. Now this product ab is not **' 
a surface, nor the representation of 
a surface. It is simply the abstract 
number 20. But this number is ex- 
actly the same as the number of 
square units contained in the rectan- 
gle whose sides are AB and B C, as 
may be seen by constructing the rec- A 
tangle AB CD. Hence the surface of 
the rectangle is measured by 20 squares described on the unit of 
length. 

This relation is universal, and we may always pass from the 
abstract thus obtained by the product of any two letters, to the 
measure of the corresponding rectangle by simply considering 
the abstract units as so many concrete or denominate units. 

In like manner, the product of three letters abc is not a solid 
obtained by multiplying lines together, which is an impossible 
operation. It is simply the product of three abstract numbers 
represented by the letters , b, and c, and is consequently an 
abstract number. But this number contains precisely as many 
units as there are solid units in the parallelopipedon whose edges 
correspond to the lines a, b, and c\ hence, we may easily pass 
from ths abstract to the concrete. Hence, if we wish to find the 
area of a rectangle whose width is 4 feet and length 6 feet, we 
simply say, 6x4=24 square feet. We pass at once from the ab- 
stract in the first member to the concrete in the second. 

It is a question whether pupils should be taught a falsehood 
in order that they may learn a truth. 

(See Bledsoc's Philosophy of Mathematics, pp. 97-106.) 



198 FINKEL'S SOLUTION BOOK. 



I. PARALLELOGRAMS. 

Prob. I. To find the area of a parallelogram ; whether it 
be a square, a rectangle, a rhomboid, or a rhombus. 



. ^4 = /X^, where ^4=area, /length, and &= 
breadth ; or, AbX&, where yl=area, =base, and tf=altitude. 

Rule Multiply the length by the breadth; or, the base by the 
altitude. 

I. What is the area of a parallelogram whose length is 15 
feet and breadth 7 feet? 

By formula, ^4=/x=lengthXbreadth=15X 7=105 sq. 
feet. 

[ 1. 15 feet=length. 
TT 1 2. 7 feet=breadth. 
H8. .'. 15X7=105 sq. ft. 
=area. 

III. .-. The area i s 105 sq. 
ft. ....... _ 

"" FIG. 4. 

Note. The base is not necessarily the side toward the ground. Thus in 
the parallelogram ABCD,BC may be considered the base, in which case, 
the altitude would be the perpendicular distance EF, between the sides BC 
and AD. If HG and B C were given, we could not find the area of the par- 
allelogram because we have not the base and altitude given. 

I. What is the area of the parallelogram ABCD, if BC is 
26 feet and EFW feet? 




By formula, ^=ax=^^X/?C=50X 26=1300 sq. ft. 

II. 26 feet=,#C= base. 
2. 50 feet=-E,F=altitude. 
3. /. 26X50=1300 sq. ft.=area. . 
III. .-. The area of ABCD=V&W sq. ft. 

I. A floor containing 132 square feet, is 11 feet wide ; what 
is its length? 



By formula, A^l^b. .-. /=^4-f-=132-f-ll=12 ft. 

II. 132 sq. ft.=area. 
2. 11 ft.=breadth. 
3. 132-:-! 1=12 ft.=length. 

III. /. The floor is 12 ft. long. 









Prob. II. The diagonal of a square being given, to find 
the area. 

Formula. Ad 2 -i-2. 

Rule. Divide the square of the diagonal by 2, and the 
quotient will be the area. 

I. What is the area of a square 
whose diagonal is 8 chains? 

By formula, ^4=<f 2 -7-2 = 8 2 -r- 
2=32 sq. chains. 

1. 8 ch.=length of diagonal= 
BD. 

II.) square described on the di- 
agonal BD. 

3. 32 sq. ch.=64 sq. ch.-f-2= 
area of the square AB CD. 

III. V.% 32 sq. ch.=the area of the FIG. 5. 

square. 

Prob. III. The area of a square being given, to find its 
diagonal. 

Formula. d=\/^A. 

. Extract the square root of double the area. 
The area of a square is 578 sq. ft. ; what is the diagonal? 
By formula, d=\/ '2A=\/2Xo7 '8= V 'l!56=34 feet. 

1. 578 sq. ft.=area of the square. 

2. 1156 sq. ft.=2x578 sq. ft.=double the area. 

3. 34 feet=\/Ti56=the diagonal. 
.-. The diagonal is 34 feet. 




II. 



III. 



Prob. IV. The diagonal of a square being given, to find 
its side. 



Formula. -S^ 

Rule. Extract the square root of one-half the square of the 
diagonal. 

I. What is the side of a square whose diagonal is 12 feet? 
By formula, S=y'p2=v'l>^=V^=6\/2=8.4852+ft 

II. 12 ft.=the diagonal. 
2. 144 sq. ft.=12 2 =square described on the diagonal. 
3. 72 sq. ft. area of square whose side is required. 
4. .-. 8.4852 ft.==6\/2==\/72= side of the square. 
III. /. The side of the square is 8.4852+ft. 



200 FiNKEL'S SOLUTION BOOK. 

Prob. V. To find the side of a square having- its area 
given. 



Rule. Extract the square root of the number denoting its 
'area. 

I. What is the side of a square field whose area is 2500 
square rods? 

By formula, S=v^=V250()=50 rods. 

( 1. 2500 sq. rd.=area of the field. 

| 2. 50 rd.=V r 2500=side of the square field. 

III. .;. The side of the field is 50 rods. 

II. TRIANGLES. 

Prob. VI. Given the base and altitude of a right-angled 
triangle, to find the hypothenuse. 



Formula. /= 



5. To the square of the base add the square of the alti- 
tude and extract the square root of the sum. 

I. In the right angled triangle A CB ', the base A C=56 and 
the altitude /?C=33 ; what is the hypothenuse? 

By formula, ^=\/rf z +* 58 =V33*+56*=\/1089+31bt)=Vi225 
=65. 

1. 56=^4 C=the base. 

2. 3136=56 2 =the square of the 

base. 

3. 33=^C=the altitude. 

4. 1089=33 8 =the square of the 

altitude. 

5. 4225=3 1364rl089=the sum of 

the squares of the base and altitude. 

6. 65 \/ 4225= the square root of the sum of the squares of 
I the base and altitude=the hypothenuse. 

III. .-. The hypothenuse=65. 

Prob. VII. To find a side, when the hypothenuse and the 
other side are given. 



II. 




Formula*.- = 

f h 



Rule. From the square of the hypothenuse subtract the square 
of the given side and extract the square root of the remainder. 



MENSURATION. 



201^ 



I. The hypothenuse of a right-angled triangle is 109, and 
the altitude 60; what is the base? 



By formula, b^/~k' 1 ^=\/109 2 00 2 =A/8281=91. 

1. 109=hypothcn.use. 

2. 11881=109 2 =square of the hypothenuse. 

3. S0=the altitude. 

II < 4. 3600=60 2 =the square of the altitude. 

5. 8281=11881 3600=difference of the squares of the 

hypothenuse and altitude. 

6. 9 1= 'V / 8281=the square root of this difTerence=the base. 

III. .'. The base is 91. 

Remark. When a=6, >h=rV/2a 2 =rt\ // 2. From this, we see that the diag- 
onal of a square is v 2 times its side. 

Prob. VIII. To find the area of a triangle, having- given 
the base and the altitude. 

Formula. A=\a x b. 

Rule. Multiply the base by the altitude and take half the 

product. 

I. What is the area of a triangle whose base is 24 feet and 
altitude 16 feet? 

By formula, A=\a^b=\ X 16x24=192 sq. ft. 

1. 24 ft.=base. 

2. 16 ft.=altitude. 

,3. 384 sq. ft.=16 X24 
IIJ duct of base and altitude. 

4. 192 sq. ft.=i of 3S4 sq. ft.== 

half the product of the base 
and the aititude=area. 

III. .. The area of the triangle is 
192 sq. ft. 




FIG. 7. 



Prob. IX. To find the area of a triangle, having given 
its three sides. 



Formula. A = 



a ) ( .v_^ ) ( x c ) , where s=^ (a-\-b-\-c). 

Add the three sides together and take half the sum; 
from the. half sum, subtract each side seperately; imdtiply the half 
sum atid the three remainders together and extract the square root 
of the product. 

* Demonstration. ^In Fi<?. 7, let A C=.b, BC-=.a, and AB=^c. In the 
right-angled triangle ADB. I3D Z AB''~~ AD*, and in the nefht-angled 
triangle CDB, BD*=BC*DC*. .'. AB*AD'*=BC' i DC' i . or r 2 



202 FINKEL'S SOLUTION BOOK 

I. What is the area of a triangle whose sides are 13, 14, and 
15, feet respectively ? 

By formula, A=\/s(sa)(sb)(s7)=*j21X (2113) X (21 
14)X(21 15) =V/21X 8X7X6 = \/7056=84 sq. ft. 

'l. 42 ft.=13 ft.+14 ft.+15 ft.=sum of the three sides. 

2. 21 ft.=4 of 42 ft.=half the sum of the three sides. 

3. 21 ft. 13 ft.=8 ft.=first remainder. 

n ,4. 21 ft 14 ft.=7 ft.=second remainder. 

^5. 21 ft. 15 ft.=6 ft.=third remainder. [mainders, 

6. 7056=21 X6X 7 X8=product of half sum and three re 

7. 84 sq. ft.= \/7056=square root of the product of the half 

sum and three remainders=the area of the triangle. 

III. .-. The area of the triangle is 84 sq. ft. 

Prob. X. To find the radius of a circle inscribed in a 
triangle. 

Formula.- R=&A-r-(a+b-\-c). 

*Rule. Divide twice the area of the triangle by the sum of 
the three sides. 

I. Find the radius of a circle inscribed in a triangle whose 
;sides are 3, 4, and 5 feet, respectively. 

1. 6 sq. ft.= V-y(j a)(s b)(s c)=area of the triangle, 
by formula, Prob. IX. 



II. 



2. 12 sq. ft. twice the area of the triangle. 



3. 12 ft.=3 ft+4 ft.+ 5 ft.=sum of the three sides. 

4. .*. 1 ft.=12-f-12=twice the area divided by the sum of 

the sides the radius of the inscribed circle. 
III. .-. The radius of the inscribed circle is 1 ft. 

AD 2 =a 2 DC 2 , whence c 2 a 2 AD 2 DC 2 . But AD 2 DC 2 =(AD 
+DC^(ADDC)=b(ADDC). .'. b(ADDC)=c 2 a 2 ,and AD DC 
=(c 2 2 )-f-. But AD-\-DC=b. .'. By adding the last two equations, we 



have 2AD= - = -- h^= - T - ; whence A>= - - - ' Since 
oo . 2o 



! , if we substitute the value of AD just found, 
( r *a*+-- 
we have 



= c! -(--^) 2 -- ^ 

-b 2 ) (2bc c 2 +a 2 b 2 )_ (b 2 -f 2bc+c 2 - 



(2Z>c+c 2 a 2 +l> 2 ) (2bc c 2 -}- a 2 b 2 )_ (b 2 + 2t>c+c 2 a 2 )[ a ^(^2bc-}-c 2 ) ] 

4b 2 4P 

[(b+c) 2 a 2 ][a 2 (b c} 2 } 



Now the 
area of ABC\A CKBD. 

- 2 ] 



%y a)(j -)(j e), where 2j=(a-f*-f-c). Q. E. D. 
*Note. For Demonstration, see any geometry. 



MENSURATION. 203 

Prob. XI. To find the radius of a circle, circumscribed 
about a triangle whose sides are given. 

abc abc 

Formula. /?= -=- . f 

4A 4 \s(s a)(s b)(s c). 

*Rule. Divide the product of the three sides by four times the 
area of the triangle. 

I. What is the radius of a circle circumscribed about a tri- 
angle whose sides are 13, 14, and 15 feet, respectively? 

1. 2730 cu. ft.=13Xl4Xl5=the product of the three sides. 

2. 84 sq. ft.= *ls(s a)(s b)(s c)=the area of the tri- 

angle, by Prob. IX. [angle. 

II.<[3. 336 sq. ft.=4x84 sq. ft.=four times the area of the tri- 
4. 8J- ft.=2730-7-336=the product of the three sides divid- 
ed by four times the area of the triangle=the radius 
of the circumscribed circle 

III, .. The radius of the circumscribed circle is 8-| ft. 

Prob. XII. To find the area of an equilateral triangle, 
having given the side. 

Fomiwlct- -A=-z V3.S" 2 , where s=side. This is what Prob. 
IX. becomes, when a=b=c. 

Rule. Multiply the square of a side by % V3,=-433013+. 

I. What is the area of an equilateral triangle whose sides 
are 20 feet? 

* Demonstration. Let ABC be any triangle, and ABCE the circumscrib- 
ed circle. Draw the diameter BE, and 
draw EC. Draw the altitude BD of the 
triangle ABC. The triangles ADB and 
BCE are similar, because both are right- 
angled triangles, and the angle *AD=the 
angle BEC. Hence, AB'-EB'- '-BD'-BC. 
Hence, AB^BCBE^BD or ac=2RX 
BD. But, in the demonstration of Prob. 

2 
IX., we found BD=-j-\/ S (8a)(8b)(sc). 

Whence 

UK 

18. 





204 FINKEL'S SOLUTION BOOK. 

By formula, A=% V3X20 2 =100 V3=173.205+sq. ft. 

II. 20 ft.=length of a side. 
2. 400 sq. ft.=20 2 =square of a side. 
3. 173.205 sq. ft.=t V3X400=.433013X400=i\ / 3 times 
the square of a side,=the area of the triangle. 

III. .'. The area of the equilateral triangle is 173.2054-sq. ft. 

Prob. XIII. The area and base of a triangle being given, 
to cut oil a triangle containing- a given area, by a line run- 
ning parallel to one of its sides. 

12' 

Formula. b'b^r-, where A=area of the given 

triangle; b^ the base of the given triangle; and A , the 
area of the portion to be cut off. 

Rule. As the area of the given triangle is to tne area of the 
triangle to be cut off, so is the square of the given base to the 
square of the required base. The square root of the result "Mill 
be the base of the required triangle. 

I. The area of the triangle ABC is 250 square chains and 
the base AB, 20 chains ; what is the base ,of the trian- 
gle, area equal to 60 sq. chains, cut off by ED parallel 
to 



II. 



By formula, ^4Z>==3 > =20 N =4V6 =-9.7979 + ch 

'1. 250 sq. ch.=area of the given triangle ABC. 

2. 60 sq. ch.=area of the 

triangle A JED. 

3. 20 ch.=base of the trian- 

gle ABC. 

4. .-. 250 sq. ch. : 60 sq. ch. 

:: 20 3 : AD*. Whence 

5. A>2= (400X60) -^-250 

=-96. 
,6. .;. ^Z^V96=9.7979+ch. FIG, 9. 

III. .-. The base ^4Z>=9.7979+ch. 

III. TRAPEZOID. 

Prob. XIV. To find the area of a trapezoid, having- given 
the parallel sides and the altitude. 




* A^(b-\-b')a, where b and b' are the parallel 
sides and , the altitude. 

Rule. Multiply half the sum of the parallel sides by the al- 
titude. 



MENSURATION. 



205 



II. 










I. What is the area of a trapezoid whose parallel sides are 
15 meters and 7 meters and altitude 6 meters? 

By formula, A=\(b+b') x=^(15+7) X6=66 m 8 . 
I. 7 m,=Z>C,the length of one 
of the parallel sides, and 

2. 15 m.=AB, the length of 

the other side. 

3. 22 m.=7 m.-(-15 m.=sum 

of the parallel sides. 

4. 11 m.=4 of 22 m.=half 

the sum of the parallel 
sides. FIG. 10. 

.5. 66 m 2 .=6xll=area of the trapezoid, ABCD. 

III. .'. The area of the trapezoid is 66 m 2 . 

IV. TRAPEZIUM AND IRREGULAR POLYGONS. 

Prob. XV. To find the area of a trapezium or any irreg- 
ular polygon. 

Rule, Divide the figure into triangles , find the area of the 
triangles and take their sum. 

I. What is the area of the trapezium ABCD, whose diag- 
onal A C is 84 feet, and the perpendiculars D JB and 
BF, 56 and 22 feet, respectively? 

1. 84 ft.=^C=Aase of the 

triangle ADC. 

2. 56 ft.=D = altitude of 

ADC. 

3. /. 2352 sq. ft.=i(^4CX 

D-')=area of the trian- 
gle AD C. 

4. 84 ft=^4C=base of the 

triangle ABC. 

5. 22 &.=!?== altitude of 

ABC. 

6. ..924sq.ft.=^(^4CX^^') FIG. 11. 

=area of the triangle ABC. 

7. 3276 sq. ft.=2352 sq. ft. +924 sq. ft.=ADC+AC= 

area of the trapezium AB CD. 

III. .-. The area of the trapezium ABCD is 3276 sq. ft. 

V. REGULAR POLYGONS. 
Prob. XVI. To find the area of a regular polygon. 

. A=\aY,pi where p is the perimeter and #, 
the apothem. 

5. Multiply the perimeter by half the apothem. 




206 



FINKEL'S SOLUTION BOOK. 



The Perimeter of any polygon is the sum of all its sides. 

The Apothem is the perpendicular drawn from the cente* to 

any side of the polygon. 

I. What is the area of a regular heptagon whose slJe is 
19.38 and apothem 20? 

1. 19.38=length of one side. 

2. 135 66=length of 7 sides=the perimeter. 

3. 20=apothem. 

4. 10=^ of 20=half the apothem. 

5. 1356.6=10Xl35.66=product of perimeter b> half the 

apothem. 

III. .-. The area of the heptagon is 1356.6. 

Prob. XVII. To find the area of a regular poiygon. when 
the side only is given. 

*Rllle. Multiply the square of the side of the -polygon by the 
number standing opposite to its name in the following table of 
areas of regular polygons whose side ^* 1 : 



Name. 


Sides. 


Multipliers. 


Triangle, 


3 


V = .4330127. 


Tetragon, or square, 


4 


1 = 1.0000000. 


Pentagon, 


5 


fVl-l |V5 = 1.7204774. 


Hexagon, 


6 


4V3 = 2.5980762. 


Heptagon, 


7 


4 cot. if 00 ^ 3.6339124. 


Octagon, 


8 


2-f 2V = 4.8284271. 


Nonagon, 


9 


f cot. 20 = 6.1818242. 


Decagon, 


10 


IV/5+2V5 = 7.6942088. 


Undecagon, 


11 


Vcot. W= 9.3656399, 


Dodecagon, 


12 


3(2+Vs) =11.1961524, 



* Demonstration. Since a regular polygon can be divided into as many 
equal isosceles triangles as it has sides, we may find the area of one trian- 
gle and multiply this area by the number of triangles, for the whole area. 
Let yl^C be one of these isosceles triangles, taken from a polygon of n 
s ; des, AB and BC the equal sides, and A C the base. The angle at the ver- 
tex 5=360-i-. ^4=K180 360-5-) C. From B let fall a perpendicu- 

/? 7~) 1 80 

lar on A C at D. Then by trigonometry, =tan (90 ). .'. BD=z 

\-A. C 



(1 80 \ 
1. The area of the triangle ABC~\ACKBD=.\AC* cot 

(_ V .'. The area of the polygons: A C 2 cot ( )r=-.y* cot ( | 
n ) 4 \^/4 \/ 

where ^=side. By placing s=l, and ^=13, 14, 15, &c., respectively, the area 
of polygons of 13, 14, 15, &c., side respectively, may be found. 




MENSURATION. 207 



Prob. XVIII. To find the side of an inscribed square of a 
triangle, having" given the base and the altitude. 

Formula. 9= r- /, where s=side, b the base, and a 
a-\-b 

the altitude. 



' 

*Rule 



* Divide the product of the base and altitude by their 
sum, 

I. What is the side of an inscribed square of a triangle 
whose base is 14 feet and altitude 8 feet? 

. ab 14X8 

By formula, s= =^=5^ feet. 

1. 8 feet=the altitude. 

2. 14 feet=the base. 

II.<j3. 112 sq. ft.=14x8=the product of the base and altitude. 
4. 22 feet=14 ft.-f 8 ft.=their sum. 
.5. SyV feet=112-7-22=the product divided by the sum 



III. .'. 5 T y ft.=the side of the inscribed square. 

VI. CIRCLE. 

Rrob. XIX. To find the diameter of a circle, having- 
given the height of an arc and a chord of half the arc. 

Formula. D=k 2 --a, in which =chord of half the 
arc and 0:=height. 

t Rule. Divide the square of the chord of half the arc by the 
height of the chord. 

* Demonstration. Let ABC be any triangle whose base is b and altitude 
a. Produce AC to H, making CHBD. At H, erect the perpendicular 
HG and make HGBD. Draw 
^4and at C, erect the perpen- 
dicular FC, and draw FK. Then 
KE=FCEN, and KN is the re- 
quired inscribed square. For, in 
the similar triangles AHG and 
ACF, we have AH:GH::AC : 
FC, or a-\-b:a::b\FC. By inver- 
sion, and then by Division, a\b 
::aFC:FC,orI:FC. In the 
similar triangles ABC and KBE, 
AC:KE::BD:BI, or BD'ACr. 
BI-KE. Whence a:b\\BI:KE. FIG ' 12 ' 

:.BI\KE\\BI\FC. :.KE=FC*n& the figure KN ^ has its sides equal 
and its angles right angles by construction. Hence, it is a square. Q.E.D. 

\Detnonstr ation. Let AB=k, the chord of half the arc ABC^ and BD 
=a, the height of the arc ABC. Draw the diameter BE and draw the 




208 FINKEL'S SOLUTION BOOK. 

I. What is the diameter of a circle of which the height of 
an arc is 5 m. and the chord of half the arc 10 m.? 

By formula, Z> = 
20 m. 



II. 



1. 10 m.=AB, the length of chord 

of half the arc. 

2. 5 m.=BD, the height of arc. 

3. 100 m. 2 square of chord. 

4. .-. 20 m.=10Oj-5=BE, the diam- 

eter of the circle. 




III. .'. The diameter of the circle is 20 

meters. FIG 13 

Prob. XX. To find the height of an arc, having- given the 
chord of the arc and the radius of the circle. 

Formula. aR *JR* c * , in which, ^?=radius and c 
=i the chord. 

*Rule. From the radius, subtract the square root of the dif- 
ference of the squares of the radius and half the chord, 

I. The chord of an arc is 12 feet and the radius of the circle 
is 10 feet. Find the height of the arc. 

By formula, aR *Jfi z c 2 =10 VlO 2 6 2 =2 ft. 

1. 10 ft.=the radius of the circle. 

2. 100 sq. ft.=square of the radius. 

3. 12 ft.=the chord. 

4. 6 ft.=half the chord. 

5. 36 sq. ft.=square of half the chord. 

6. 8 ft.=\/ioo 36 = square root of the difference of the 

squares of the radius and half the chord. 

7. .'. 10 ft 8 ft.=2 ft.=height of the arc. 

III. .-. The height of the chord is 2 feet. 

Prob. XXI. To find the chord of half the arc, having- 
given the chord and height of an are. 

Formula. > 



n. 



radius A O. The triangles ADB and BAE are similar, because their an- 
gles are equal. Hence, BE\AB\'.AB:BD, or BE:k::k:a. Whence BE 
z=Z>r= 2 -5-a. Q.E.D. 

N. B. (1) If a and D are given, =y 'Z>X; (2) if D and k are given 



* Demonstration. In Fig. 13, we have BD=BODO. But Z>O= 
2 >A 2 =tf[jR 2 c?]. /. aR^R 2 c 2 . If a and 7? are given, (1) 
a) 2 ] = 2>J(2ajR a 2 ); if a and 2c are given, 






MENSURATION. 



209 



*Rule. Take the square root of the sum of the squares of the 
height of arc and half the chord. 

I. Given the chord=48. the height=10, find the chord of 
half the arc. 

1. 48=the chord. 

2. 576=4 of 48 2 =square of half the chord. 

3. 1Q= height of chord. 

4. 100 v quare of height of chord. 

5. 676=576+100=sum of square of half of chord and 

height. 

6. 26=^676 square ro t f sum f square of half of chord 

and height. 
III. ^ /. The chord of half the arc is 26. 

Prob. XXII. To find the chord of half an arc, having: 
given the chord of the arc and the radius of the circle. 






Formula. k=%R* ^?\/4/? 2 4c 2 . 

. Multiply the radius by the square root of the differ- 
ence of the squares of twice the radius and the chord; subtract 
this product from twice the square of the radius and extract the 
square root of the difference. 

I. Given the chord of an arc=6 and the radius of the circle 
=5, find the chord of half the arc. 

By 



1. 5=the radius of the circle. 

2. 10=twice the radius of the circle. 

3. 100=square of twice the radius. 

4. 6=chord of the arc. 

5. 36=square of the chord. 

6. 100 36=64=difFerence of squares of twice the radius 

and the chord. 

7. 8=\/0-i~ sc l uare r ot of the above difference. 

8. 40=5X8 the product of the above square root and the 

radius. 

9. 50=2x5 2 =twice the square of the radius. 

10. \/oO 40=V'lO=chord of half the arc. 
III. .'. The chord of half the arc is \/i(j. 



II. 



* Demonstration. In Fig. 13, we have AB= A 
\/ 2 -)-c 2 . .-.=V<7 2 -f-c 2 . If and 2c are given, (1)#=:^ 

^Demonstration. From Prob. XXI., we have &=^a z -\-c 2 . 
XX. we have a=j 



if k and a are 

From Prob. 
c 2 . Substi- 



tuting this value of a z in the above equation, =V2JR 2 



210 



FINKEL'S SOLUTION BOOK. 



Prob. XXIII. To find the side of a circuinscribed polygon, 

he circle and a side of a simi- 



having- given the radius of the 
lar inscribed polygon. 



. K'-=. 



in which K' is the side of 



V4/? 2 A' 2 

the circumscribed polygon and TTthe side of a similar 
inscribed polygon. 

Divide twice the product of the side of the inscribed 
'polygon and radius by the square root of the difference of the 
squares of twice the radius and the side of the inscribed polygon. 
I. When /?=!, find one side of a regular circumscribed do- 
decagon. 

^KR ^K 

By formula, K'= .. . The formula does 

^RiK* *JK* 

not lead to a direct result, since K is not given. But 
by the formula of Prob. XXI., if k is replaced by K 
we have KVz V< -1 f r 2c=l, since it is the side 
of a regular inscribed hexagon, and K=V% \/3> s i nce 
2c is a side of a regular inscribed dodecagon. 




, 

~ 



VII. 



^ 



RECTIFICATION OF PLANE CURVES AND 
QUADRATURES OF PLANE SURFACES. 



1. To ^Rectify a Curve is to find its length. The term 
arises from the conception that a right line is to be found which 
has the same length. 

2. The Quadrature of a surface is finding its area. The 
term arises from the conception that we find a square whose 
area is equal to the area of the required surface. 

The formula for the rectification of plane curves is 

dx, when the curve is re- 



ferred to rectangular co-ordinates. 



curve is referred to polar co-ordinates. 



* 2dz 2 



are formulae for the rectifica- 
tion of curves of double cur- 
vature, when referred to rec- 
tangular co-ordinates*. 



MENSURATION. 211 



are formulae for the 
rectification of 
j curves of double 
' r curvature, referred 
to polar co-ordi- 
nates. 

A= Cydx or Cxdy is the formula for the quadrature of any 
plane surface referred to rectangular co-ordinates. 

A.=? C^r z dO is the formula for the quadrature of plane surfaces, 
referred to polar co-ordinates. 

3. A Surface Of Revolution is the surface generated 
by a line (right or curved) revolving around a fixed right line as 
an axis, so that sections of the volume generated, made by a plane 
perpendicular to the axis are circles. 



/ y ->J ! 1-f- 1 { dx is the formula for a surface of revo- 
J J N r ' \dx\ ) 

lution, referred to rectangular co-ordinates. 
=27T fyds=&7t Cr sin 6\ r*+\-^\ dQ is the formula for 



a surface of revolution, referred to polar co-ordinates. 

K=TT fy*dx or x z dy is the formula for the volume of a solid 
of revolution referred to rectangular co-ordinates. 

V= C C Cdxdydz and V= C Czdxdy are formulae for the 
cubature of solids, requiring triple and double integration. 

F= f CzrdBdr and V=CJJr 2 sin SdcpdQdr are the formulas 
for cubature of solids referred to polar co-ordinates. From 
the equation to the surface of the solid, z must be expressed 
as a function of r and 6. 

__j^>2 i s the rectangular equation of a circle referred to 
the center. 

y 2 ==2-/?# x z is the rectangular equation of a circle referred to 
the left hand vertex as origin of co-ordinates. 

r=%R cos.# is the equation of the circle referred to polar co-ordi- 
nates. 

Prob. XXIV. To find the circumference of a circle, the 
radius being given. 



212 FINKEL'S SOLUTION BOOK. 




" Rdx 1 1.3 1.3.5 



=4ff Xl-570796+=3.141592x2.ff=2;r J ff, in which 

z..o.o. 

7T=3.141592+. Since the diameter is twice the radius, we have 
27rft=7tD, in which D is the diameter. .*. C,=%7tR=7tD. 

C C 

.'. (1) R=^, (2)Z>= , where C is the circumference. 

TC 7i 

Rule. Multiply twice the radius or the diameter by 3.141592. 

I. What is the circumference of a circle whose radius is 17 
rods ? 

By formula, C=2^/?=3.141592 X 34 rods = 106.814128 
rods. 

1. 17 rods=the radius. 

2. 34 rods=2Xl7 rods=the diameter. 

3. 106.814128 rods=3.141592X34rods=the circumference. 

III. .-. The circumference is 106.814128 rods. 

Note. The ratio of the circumference to the diameter can not be exactly 
ascertained. An untold amount of mental energy has been expended upon 
this problem; but all attempts to find an exact ratio have ended in utter 
failure. Many persons not noted along any other line, claimed to have 
found this clarem impossibilitibus by which they have unlocked pll the diffi- 
culties that have encumbered the quadrature of the circle for more than two 
thousand years. The Quadrature of the Circle is to find a square whose area 
shall be exactly equal to that of the circle. This can not be done, since the 
ratio of the circumference to the diameter can not be exactly ascertained. 
Persons claiming to have held communion with the "gods" and extorted 
from them the exact ratio are ranked by mathematicians in the same class 
with the inventors of Perpetual Motion and the discoverers of the Elixir of 
Life, Alkahest, the Fountain of Perpetual Youth, and the Philosopher's 
Stone. Lambert, an Alsacian mathematician, proved, in 1761, that this ratio 
is incommensurable. In 1881, Lindemann, a German mathematician, dem- 
onstrated that this ratio is transcendental, and that the quadrature of the 
circle by means of the ruler and compass only, ort>y means of any algebraic 
curve, is impossible. Its value has been computed to several hundred deci- 
mal places. Archimedes, in 287 B. C., found it to be between 3f? and 3f ; 
Metius, in 1640, gave a nearer approximation in the fraction ff |; and, in 1873, 
Mr. W. Shank presented to the Roj^al Society of London a computation ex- 
tending the decimal to 707 places. The following is its value to 600 deci- 
mal places: 

3. 141, 592, 653, 589, 793,238,462,643,383, 279, 502,884, 197, 169, 
399,375,105,820,974,944,592,307,816,406,286,208,998,628,034,825, 
342,117,067,982.148,086,513,282,306,647,093,844,609,550,582,231, 
725,359,408,128.481,117,450-284,102.701,938,521,105,559,644,622, 
948 ,954,930,381 ,964,428,810,975 ,665,933,446, 128 ,475 ,648 ,233,786, 



MENSURATION. 213 

783,165,271,201,909,145,648,566,923,460,348,610,454,326,648,213, 
393,607,260,249,141,273,724,587,006,606,315,588,174,881,520,920, 
962,829,254,091,715,364,367,892,590,360,011,330530,548,820,466, 
521,384,146,951,941,511,609,433,057,270,365,759,591,953,092,186, 
117,381,932,611,793,105,118,548,074,462,379,834,749,567,351,885, 
752,724,891,227,938,183,011,949,129,833.673,362,441,936,643,086, 
021,395,016,092,448,077,230,943,628,553,096,620,275,569,397,986, 
950,222,474,996,206,074,970,304,123,669+. 



Bernoulli's Formula. 



"'' Wallis ' s Formula > 1666. 



i7T=l+l_. ....... Sylvester's Formula, 1869. 

1+1.2 
1+2.3^ 
1+3.4 

M-L 

1+4.5 



2+ ...... Buckner's Formula. 

The Greek letter n, was first used by Euler, to designate the 
ratio of the circumference to the diameter. 

Prob. XXV. To find the length of any arc of a circle, 
having- given the chord of the arc and the height of the arc, 
i. e., the versed sine of half the arc. 



(a). 

Rdx 



2)3 
2 \3 ' 



the arc and c half the chord of the arc. 



*Note. This series was discovered by Bernoulli, but he acknowledged 
his inability to sum it. Euler found the result to be ^TT*. For an inter- 
esting discussion of the various formulae for TT, see Squaring- the Circle, 
Britannica Encyclopedia 



214 FINKEL'S SOLUTION BOOK. 

(b.) Formula. s=a.rc=^(8& 0)*, where a is the chord of 
the whole arc and b the chord of half the arc. 

Rule from (b) . From eight times the chord of half the arc sub- 
tract the chord of the whole arc ; one-third of the remainder will 
be the length of the arc, approximately. 

I. Find the length of the arc whose chord is 517638 feet 
and whose half chord is 261053.6 feet. 

By formula (b), s=%(8&a)=% (8X261053.6517638)= 
52359.88 feet. 

f 1. 261053.6 feet=length of chord of half arc. 

2. 2088428.8 feet=8X261053.6 feet=eight times the 

length of chord of half arc. 

3. 517638 feet=rlength of chord of whole arc. 



II. < 



4. 1570790.8 feet=2088428.8 feet 517638 feet=differ- 



ence between eight times chord of half arc and 
chord of whole arc. 

5. 52359.69 feet=J of 1570790.8 feet=length of arc, 
nearly. 

III. .-. The length of the arc is 52359.69 feet. 

Note. This important approximation is due to Huygens, ( he wrote his 
name Hugens. It is also sometimes spelled Huyghens), a Danish mathe- 
matician, born at the Hague, April 14, 1629, and died in the same town in 
1695. For a brief biography of this noted mathematician, see Ball's 
A Short History of Mathematics, pp. 302-306. 

The following is Newton's demonstration : 

Let R be the radius of the circle, L the length of the arc, A the chord 
of the arc, and B the chord of half the arc. 

A L B L 

Then -^=2 sm.^, -^=2 sm.^. 

X X Z X*> 

Since, sin. #=- gj-j-yj- etc. (See Bowser's Treatise on Trigonom- 
etry, or any other good work on the subject), we have 

and 



4^=2 ( 2*-V2.#) +V2^/ -etc. ) 

* V~ ~~3T~ ~~5T~ / 

=<?- 




. * . A L A? > 2 + o4^i >4 e tc., and 
8#=4Z, 



, nearly,=A nearly. 

In the problem proposed, the radius is 100000 feet and the arc is 30. 
Using 7r=3 . 1415926, =52359 . 88 feet. . . The result by the formula lacks 
only about 2 inches of being the same. 



MENSURATION. 215- 



(c.) *orm^-^=*rc=a( x 

This formula is a very close approximation to the true length 
of the arc when a and c are small. The first formula may be ex- 
tended to any desired degree of accuracy. 

Rule from (c). Divide 10 times the square of the height of 
the arc by 15 times the square of the chord and 33 times the height 
of the chord; multiply this quotient increased by 1, by 2 times the 
square root of the sum of the squares of the height and half the 
chord. 

I. The chord of an arc is 25, and versed-sine 15, required the 
length of the arc. 

02_|_ c Br- c 2__02 ( C 2_ a 2}3 3 

By formula (a), arc= -- [^-^ [^^ \ +^ 

5 (c*a*Y & -]_15 2 -t-25 2 r25 2 -15 2 1 
l2|/i a +c 2 J ' ' C J~ 2X15 Ll5 2 +25 2 ~ i ~6 X 

3 r25 2 -15 2 ]5 5 r26-15n 7 -] 
15 2 +25 2 J ^ 15 2 +25 2 J + n &C 



+ft. 

1. 25 ft.=length of the chord. 

2. 15 ft. height of the arc, or the versed-sine. 

3. 2250 sq. ft.=10 times 15 2 =10 times the square of the 

height of the arc. [chord. 

4. 9375 sq. ft.=15 times 25 2 =15 times the square of the 

5. 7425 sq. ft=33 times 15 2 =33 times the square of the 

height of arc. 
!!.<{ 6. 17HOO sq. ft.=7425 sq. ft.+9375 sq. ft. 

7. ^=^2250-7-17800=10 times 15 2 -r-(15 times 25 2 +33 

times 15 s ). 

8. 14^==4==1 + 10 times 15 2 -j-(15 times 25 2 +33 

times 15 2 ). 

9. 381J sq. ft.=15 2 +(12j) 2 . 

10. 53.58 ft.=-|fixVl5 2 +(12i) 2 =|fiX|V61^=len g th of 

arc, nearly. 
III. .'. 53.58 ft.=length of the arc. 

Prob. XXVI. To find the area of a circle having given the 
radius, diameter, or circumference. 



Formula. 




C, when the ra- 



216 FINKEL'S SOLUTION BOOK. 

dius and circumference are given. /. (1) R=*jA-r-rt, (2) D= 
=2 R=2V.4-v-;r, and (3) C=\ / 4^4= 



Rule I. The area of a circle equals the square of the radius 
multiplied by 3.14-1592; or (2) the square of the diameter multi- 
plied by .785398; or (3) the square of the circumference multiplied 
by .07958/ or (4) the circumference multiplied by \ of the diame- 
ter; or (5) the circumference multiplied by \ of the radius. 

Rule II. Having given the area. (1) To find the radius: 
Divide the area by 3.14159%, and extract the square root of the 
quotient. (2) To find the diameter : Divide trie area by 3.141592 
and multiply the square root of the quotient by 2. (3) To find 
the circumference: Multiply the area by 3.141592 and multiply 
the square root of the product by 2. 

I. What is the area of a circle whose radius is 7 feet? 

By formula, ^=7T.ff 2 =3.141592x7 2 =153.93804+ sq. ft. 



1. 7 ft.=the radius. 
5. 49 sq. ft.=7 2 =square of the radius. 

153.93804 sq. ft.=3. 141592x49 sq. ft.=area of the circle. 

III. .'. 153.93804 sq. ft.=area of the circle. 



IL/2. 
13. 



I. What is the area of a circle whose diameter is 4 rods ? 

By formula, ^=i7rZ> 2 =4x3.141592x4 2 =12.566368 sq. ft. 

II. 4 ft.=the diameter. 
2. 16 sq. ft.=square of the diameter. 
3. 12.566368 sq. ft.=ix3.141592x4 2 =.785398Xl6 sq. ft. 
=area of the circle. 

III. .-. 12.566368 sq. ft.^area of the circle. 
I. What is the area of a circle whose circumference is 5 meters ? 



By formula, ^4=.=^=1.989 m 2 . 

{1. 5 m.=the circumference. 
2. 25 m. 2 =the square of the circumference. 
3. 1.989 m. 2 =.07958x25 m. 2 =the area of the circle. 
III. /. 1.989 m. 2 =the area of the circle. 

Remark. We might have found the radius by formula (1) under Prob. 
XXIV and then applied the first of Rule I. above. We might have found the 
radius by formula (1) of Prob. XXIV and then applied (5) of Rule I. above. 

I. What is the circumference of a circle whose area is 10 A.? 



MENSURATION. 



217 



By formula (3), C=2*/7rA=M3. 141592 X 1600=80V?r80X 
1.7724539141.796312 rods. 

1. 10 A.1600 sq. rds.=the area of the circle. 

2. 1600H-7T=the square of the radius. 

II.< 



3. .-. 40 V^the radius. 



71 



80 ,_ 40 - 

~7t ^ 7f==7r ^ times ~V7T=the diameter. 

40 



5. SQ\x=7TX VTT 141.796312 rods=the circumference. 
III. .'. 141.796312 rods=the circumference of the circle. 

I. With what length of rope must a horse be tied to a stake 
so that he can graze over one acre of grass and no more? 

By formula (1), 7?=Vyl~^=Vl60-r-7r=4 > | =7.1364+ rd. 

1. 1 A.=160 sq. rd.=area of the circle over which the 

horse can graze. 
IL )2. 160-7- 7r==square of the radius. 

3. Vl60-r-7T=4VlO-7~7T=7.1364 rd.=radius or length of rope. 
III. .-. 7.1364 rd.=length of the rope. 

Prob. XXVII. To find the area of a Sector, or that part of 
a circle which is hounded by any two radii and their included 
arc, having griven the chord of the arc and the height of the 
arc. 



- i* c z a* ( (a z +c*} 2 (c* a*}*} 

sm 7r-^l^ ~ t 




+ 

:.j in which c is half the chord of 
arc and a the height of arc. 

Demonstration. Let AB=x, BD=y, and ft=A>=thc radius of the cir- 
cle. Then x 2 -}-y 2 =K 2 , the equation of the circle referred to the center. 
Now A 2 Cydx; buty=(7? 2 x 2 ) l A, from the equation of the circle. 

/. A=.2 C(R 2 x*}Xdx=x (R 2 x 2 }X-\-R* sin~ 1 -^. But x=R a and y=c. 



Hence ^=(7? a)[R* (/? a) 2 ]+7? 2 sin~~. Ikit,from(2) Prob. XX, 



218 



FINKEL'S SOLUTION BOOK. 



Rule. (1) Find the length of the arc by Problem XXV, 
and then multiply the arc by half the radius which may be found 
by Problem XX, in 'which c and a are known and R is the un- 
known quantity. 

(2) If the arc is given in degrees, take such a part of the 
whole area of the circle as the number of degrees in the arc is 
ef360*. 

I. Find the area of the sector, the chord of whose arc is 40 
feet, and the versed-sine of half the arc 15 feet. 



By formula, 



20 2 



c 20 2 15 2 



1 20215 



II. 



III. 



20 2 +15 2 
1. 53.58 ft.= 



,3 , 1 _ f20152'|6 
J +1.2.3.4.5 l20+15J " 



length of the arc, by (), Prob. XXV. 






2. 20f ft.= 



# 2 I C 2 








=radius of the circle, by solving th 



formula of Prob. XX with respect to R. 
3. .'. 558.125 sq. ft.=-J (20-|x53.58)=area of the sector. 

.'. 558.125 sq. ft.=the area of the sector. 




Trigonometry, we have sin~ l O=0 -^ 



But ' fron 



_#s Ci Hence, A=z 



2a 



In this formula, 



c2a*) * c r 2 a 2 __ 1_ [ r 2 a* 3 . 1 (c 2 a^ <t ^ 

2a } i c^+a* 1.2.3 U 2 + 2 J 1-2.3.4.5 U 2 + 2 J " \ 



c is the area of the tri- 



angle DBA. For *:=/? = 



^a-LcZ c*a* 

-JT a n : 

2a 2a 



the altitude and c is half the base of the triangle 

( C 2 _ a 2} 

.: - - \c = the area of the triangle DBA. 

Therefore, if we subtract the area of the tri- 
angle DBA from the area of the sector,, we 
shall have the area of the segment DEC. Hence, 

( c z a z\ 2 

-the area of the segment DEC is - 

2a 




be carried to any desired degree of accuracy. 



FIG 74. 



MENSURATION. 219 

I. What is the area of a sector \vhose arc is40and the radius 
of the circle 9 feet? 

1. 9 ft.=radius of the circle. 

2. 7n?? 2 =7z9 2 =area of the whole circle. 
IU3. 40=length of the arc. 

4. 40Hrof360. 

5. 7r9=4 of 7r9 2 =28.274328 sq. ft.=area of the sector. 
III. ' .-. The area of the sector is 28.274328 sq. ft. 

Prob. XXVIII. To find the area of the segment of a circle, 
having- given the chord of the arc and the height of the 
segment, i. e., the versed-sine of half the arc. 



( c z_a z } 1 (c* a*} 

__ 



, 

- &c - 



. Divide the cube of the height by twice the base and in- 
crease the quotient by two-thirds of the -product of the height and 
base. 

I. What is the area of a segment whose base is 2 feet and al- 
titude 1 foot? 

By formula (*), A^ 



* Demonstration. In the last figure, let jB C=a=altitude of the segment and 
DE2c=\hQ base of the segment. Then BD Z = BCXSF=a (2R a)c z . 

Whence R= C J~ rt , and AD=R a= C ' a ~ C a . Z?C= v/ZM^rz^ 
2a 2a 2 

/? 77 r 

By Trigonometry, r=sin/Z>^4C, or rrr sin/ DAC. Now 27r^?r= 
yl .Zx oi 



180 , 180 . 

-- Therefore, R: arcDC:: - : ?= - - X -- Let s = arc 

7T 7T R 7T 



DCE. Then the /DAC=-=. .'. --^sin-- . In like manner, from 

ft Zff JTL Zfi 

the right angled triangle ^Z>C,-,=rsin/C^Z?, or since the 



^ / CAD, ^5 sin . Now since the sine of any angle fczfJ 
., the above equation becomes 
&c (1), and 




ff 



&c (2). Multiplying equation (2) 

by 8 and subtract equation (1) in order to eliminate the term containing 
s*, we have approximately ,^^- C =r^| % fr-^-^Tl (nTs) + &c. Omitting 



the negative quantity, since it is very small in comparison with s and be- 
cause it is still more diminished by a succeeding positive quantity, we have 



II. 



FINKEL'S SOLUTION BOOK. 

1. 1 ft.=altitude of the segment. 

2. 2 ft.=base of the segment. 

3. 4 ft.=twice the base of the segment, 

4. 1 cu. ft cube of the height of the segment. 

5. i sq. ft.=l-7-4=quotient of the cube of the height and 



twice the base. 

6. 2 sq. ft.=2 X l=product of the height and base. 

7. lij- sq. ft.=f- of the product of the height and base. 

1-J- sq. ft.-f-^ sq. ft.=l T 7 ^sq. ft.==area of the segment. 
III. .. The area of the segment is 1 T 7 ^- sq. ft. 

Prob. XXIX. To find the area of a circular zone, or the 
space included between any two parallel chords and their 
intercepted arcs. 

, \ A c*a 2 *c*a 2 1 
.-a A= 



) __ (c'*a 

" ' 1 " M 




1.2.3 



Rule. Find the area of each segment by Prob. XXVIII., and 
take the difference between them, if both chords are on the same 
side of the center; if on opposite sides of the center, subtract the 
sum of the areas of the segments from the 'whole area of the circle^ 

I. What is the area of a zone, one side of which is 96, and 
the other 60, and the distance between them 25? 

Let ^4^=60=2^, CY?=96=2c, and HK=2b=h. Then AH 
=30=^ and CK=S=c. Let OA=R. Then 

But 




ne. is. 



3 - g =%(4\/c 2 +a 2 a). This is the approximate 

length of an arc in terms of its height and base. Now the area of the seg- 
ment >CJS=% A CXarcDCE area of the triangle DEAy z RxS% AB 

X 




Q. E. D. 



MENSURATION. 221 



h* c / ' a ') s . In like manner, LKa=R i/J?z _ c*,=i 




.-. By formula (), ^= 



X 



(48 8 25 2 3Q 2 )=radius of the circle. 

2. OK=>lR*c* =V50 3 48 g =14. 

3. .*. XA====50 14=36 altitude of segment 

4. O H v ' R ^ c' 2 " V^5Q 2 30 * 40. 

5. .-. Z^T= / =50 40=10=altitude of the segment 

36 3 

+f(96X36)=2547=area of segment CDBLA. 



10 3 

7. x 4-f(60X10)=408i=area of the segment 

8. .'.2547 408i==2138f=area of the zone CDBA. 
III. .'. 2138f=area of the zone ABDC. 

Note. This result is only approximately correct. The radius of the cir- 
cle may be found by the following rule: 

Subtract half the difference between the two half chords from the greater 
half -chord, multiply the remainder by said difference, divide the product by 
the width of the zone, and add the quotient to half the width. To the square 
of this sum add the square of the less half chord, and take the square root of 
the sum. 

This rule is derived from the formula in the above solution, in which 



222 FINKEL'S SOLUTION BOOK. 

Prob. XXX. To find the area of a circular ring, or the 
space included between the circumference of two concen- 
tric circles. 

Formula,* (.) A=n (7? 2 r 2 ), in which R and r are 
the radii of the circles. 

(b.) *A=%7rc*, in which c is a chord of the larger circle 
tangent to the smaller circle. 

I. Required the area of a ring the radii of whose bounding 
circles are 9 and 7 respectively. 

By formula (), A = 7t(R* r*) = 7r(9 2 7 2 )=32?r= 

100.530944. 
'1. 9=/?=:radius of the larger circle, and 

2. 7=7'=radius of the smaller circle. 

3. 7r9 2 =7r^? 3 =area of larger circle, and 

4. 7r7 2 =7T?' 2 =area of smaller circle. 

5. .'. 7T9 3 7r7 2 = 7r(9 s 7 2 ) = 327T= 

100.530944=area of the ring. 
III. /. 100.530944=the area of the ring. 

* Demonstration. Let ABC be the chord of the 
large circle, which is tangent to the smaller 
circle, and let AC=c. Then BC=%c= 



II.< 




and Jf7rc 2 r=/r(^?2_ r 2) But K(R* r 2 ) is the dif- 

ference of the areas of the two circles or the area 

of the ring, *. ^7rc 2- ^:the area of the ring. 

Q ). FIG - 16 ' 

Prob. XXXI. To find the areas of circular lunes, or the 
spaces between the intersecting* arcs of two eccentric circles. 



Formula. ^= 



* Find the area of the two segments of 'which the lunes 
are formed^ and their difference 'will be the area required. 

I. The chord AB is 20, and the height DC is 10, and DE 2; 
find the area of the lune AEB C. 




+</*. V . 

If now we find the altitudes of the two segments and then find the length 
of the arcs of the segments by formula (), Prob. XXV, and then find the 
area of the sectors by multiplying the length of the arcs by half the radius, 
from the areas of the sectors subtract the triangles formed by the radii of 
the circles and the chord of the arcs, we shall then have the area of the two 
segments. Taking their difference, we shall have for the area of the zone 
2136*75, which is a nearer approximation to the true area. 



MENSURATION. 



223 



By formula, A= 



II. 




2. DE = height of segment 

3. DC= height of segment 

area of the segment A CBD. 



FIG. n. 

area of the segment AEBD. 
III. .-. 158^ 26ff=131 T V=area of the lime A CBE. 




VIII. CONIC SECTIONS. 



1. The ConiG Sections are such plane figures as are 
formed by the cutting of a cone. 

2. If a cone be cut through the vertex, by a plane which also 
cuts the base, the sections will be a triangle. 

3. If a right cone be cut in two parts, by a plane parallel to 
the base, the section will be a circle. 

4. If a cone be cut by a plane which passes through its two 
slant sides in an oblique direction, the section will be an ellipse. 

5. The Transverse Amis of an ellipse is its longest 
diameter. 

6. The Conjugate Amis of an ellipse is its shortest 
diameter. 

7. An Ordinate is a right line drawn from any point of 
the curve, perpendicular to either of the diameters. 

8. An Abscissa is that part of the diameter which is con- 
tained between the vertex and the ordinate. 

9. A Parabola is a section formed by passing a plane 
through a cone parallel to either of its slant sides. 

1C. The Axis of a parabola is a right line drawn from the 
vertex, so as to divide the figure into two equal parts. 

11. Tile Ordinate is a right line drawn from any point 
in the curve perpendicular to the axis. 



224 FINKEL'S SOLUTION BOOK. 

12. The Abscissa is that part of the axis which is con- 
tained between the vertex and the ordinate. 

13. An Hyperbola is a section formed by passing a 
plane through a cone in a direction to make an angle at the base 
greater than that made by the slant height. It will thus pass 
through the symmetrical opposite cone. 

14. The Transverse Diameter of an hyperbola, is that 
part of the axis intercepted between the two opposite cones. 

15. The Conjugate Diameter is a line drawn through 
the center perpendicular to the transverse diameter 

16. An Ordinate is a line drawn from any point in the 
curve perpendicular to the axis. 

17. The Abscissa is the part of the axis intercepted be- 
tween that ordinate and the vertex. 

1. ELLIPSE. 

a 2 y 2 -^-& 2 x 2 =a 2 l> 2 is the equation of an ellipse referred to the 
center. 

32 
y 2 = (%ax AT 2 ) is the equation of the ellipse referred to left 

hand vertex. 

In these equations, a is the semi-transverse diameter and b the 
semi-conjugate diameter \y is any ordinate and x is the corres- 
ponding abscissa. When any three of these quantities are given 
the fourth may be found by solving either of the above equations 
with reference to the required quantity. 

p=^ is the polar equation referred to the centre, and 

1 e cos 6 

a t i e z \ 

p= -r n is the polar equation referred to the left hand 

l-\-e cos 6 

vertex. 

Prob. XXXII. To find the circumference of an ellipse, the 
transverse and conjug-ate diameters being- known. 



Formula. cir.= C= 




MENSURATION. 



1-3 - y-\ 3* 6 (50 2 r3a 2 f0 2 . , t x * Ari - 5 1 
/r2 _ v2 I __ - J - 1 - I sin -- \a 2 x 2 \ _ 
* J 2.4.6 5 ( 6 L 2 [2 a 2* } 

a A f na e * fa na\ 
=4 VTA2 ' T) - 




4 V2'2 ~ * 



r ^2 
| !___ 



Rule. Multiply the square root of half the sum of the squares 
of the two diameters by 8.141592, and the product will be the cir- 
cumference, nearly. 

I. What is the circumference of an ellipse whose axes are 24 
and 18 feet respectively ? 



By formula, Ctr.= C=27T X 12 

-JV &c. =27rx 12 X- 87947=66.31056 ft., nearly. 



1. 576 sq. ft.=24 2 =square of the transverse diameter. 

2. 324 sq. ft.=18 2 =square of the conjugate diameter. 

3. 900 sq. ft. sum of the squares of the diameters. 

4. 450 sq. ft.=half the sum of the squares of the diameters. 
' 5. 15V2ft.=V450=square root of half the sum of the squares 

of the diameters. 

6. 7rl5\/2 ft.= 66.6434 ft., nearly, =the circumference of 
the ellipse. 

III. .-. The circumference of the ellipse is 66.6434 ft. nearly, by 
the rule. 

Prob. XXXIII. To find the length of any arc of an ellipse, 
having given the ordinate, abscissa, and either of the diam- 
eters. 

Formula. s-=Z[%7ta\l (|) 2 ^ (|-i) 2 4 (i-M) 2 

J. o 

<? 6 ) i .x e* fa* . x x } e 4 f~3a 2 

&c I I a sin 1 sin ^a x 2 - -\ T 

5 j J a 2a[2 a V } 2.4 3 L 4 

C * 2 v y . V 3 ^ , ^-^ J 

sin" 1 Wa 2 x 2 -~\a 2 x* &c., in which x is the ab- 

[2 a 2 4 _JJ 

\a*b* 

scissa: a the semi-transverse diameter ; and =->J = =the ec- 

a 

ccntricity of the ellipse. 



226 FJNKEL'S SOLUTION BOOK. 

Rule. Find the length of the quadrant CB by Prob. XXXI 
and CF by substituting the value of x in the above series. Twice 
the difference between these arcs will give the length of the arc 
FBG. 

I. What is thelengthofthearc7^G,if OE=x=$,EF=y=%, 
and 0C=3=10? 

Since 2 jx 2 +3 2 * 2 =# 2 2 , we find, by substituting the values of 
x,y, and b, 0=15. Then by the formula, FBG=s=& %na j 1 



x e*(a z t x *r - -) 

a sm- 1 - sin" 1 -- 7 Va z x 2 \ 

a 2a (2 a 2 J 




FIG. 18. 

/io\a v ' / i o \ O "" O / C\ t * ** + u 

(I4) 2 ^ (M'f) K &c * ( 2 )15sm - 1 



__,! / 

-9^J - &c. J =,rl5x.815-2 j g* 




.15 2 f!5 2 . _. 9 
IB 



Prob. XXXIV. To find the area of an ellipse, the trans- 
verse and conjugate diameters being- given. 



Formula. A=4<tx=4 



2 x*)dx=nab, in 

which a and b are the semi-transverse, and semi-conjugate diame- 
ters. 



. Multiply the product of the semi -diameters by n=^ 
3.141592, or multiply the product of the diameters by \7t=.785398. 
I. What is the area of an ellipse whose traverse diameter is 70 
feet and conjugate diameter 50 feet? 

By formula, ^=^^=^35X25=2748.893 sq. ft. 

rl. 35 ft=| of 70 ft=length of the semi-transverse diameter. 
II.< 2. 25 ft=-^ of 50 ft.=length of the semi-conjugate diameter.. 

13. /. 2748.893 sq. ft.=7rx35x25=the area ofthe ellipse. 
III. .-. The area of the ellipse is 2748.893 sq. ft. 



NOTE. 7ra3 = ^7r 2 .7r^2. .. The area of an ellipse is a mean propor- 
tional between the circumscribed and inscribed circles. 



MENSURATION. 



227 



Prob. XXXV. To find the area of an elliptic segment, hav- 
ing- given the base of the segment, its height, and either di- 
ameter of the ellipse, the base being parallel to either diam- 
eter. 

Formulae. (a)A=J ydx, or -J xdy- Cx(a* 



'in (ij '-2 (&) ' 



(b) A= y dx==x(a z -- **)i+a 2 sin' 1 



The former formula of (a) gives the area of a segment whose base 
is parallel to the conjugate diameter and the latter the area of a 
segment whose base is parallel to the transverse diameter. 

Rule. Find the area of the corresponding segment of the 
circle described upon the same axis to 'which the base of the seg- 
ment ts perpendicular. Then this axis is to the other axis as the 
area of the circular segment is to the area of the elleptic segment. 

2, PARABOLA. 

y vZpx is the rectangular equation of the parabola. 

p 

p== - - 7, is the polar equation of the parabola. 
1 cos 6 

In the rectangular equation, HGy, the ordinate; AG=x t the 
abscissa; A.F=.AE-=\p. If any two of these are given the re- 
maining one may be found from the equation, p is a constant 
quantity. 

Prob. XXXVI. To find the length of any arc of a parabola 
cut off' by a double ordinate. 



Formula. s=2j<Sdy*+dx*== C(p 2 -f 




228 



FIXKEL'S SOLUTION BOOK. 



Rule. When the abscissa is less than half the ordinate: To 
the square of the ordinate add of the square of the abscissa and 
twice the square root of the sum will be the length of the arc. 

I. What is the length of the arc KAff, if A G is 2 and GH 6 ? 



By formula, j?== 



flag 



Since j>/ 2 =2/*, we have 




length of the arc, nearly. 
"1. 2=A G=the abscissa. 
2. 6= GH=\hQ ordinate. 



II. 



3. 36=the square of the ordinate. 



4. y f of 2 2 =4 of the square of the abscissa. 

5. 2\f(y+36)==12.858==the length of the arc, nearly. 
III. .'. 12.858=length of the arc, nearly. 

Prob. XXXyil. To find the area of a parabola, the base 
and height being- given. 

Formula. A=2 

i. e., the area of parabola HKA is f of the circumscribed rectangle. 

Rule. Multiply the base by the height and f of the product 
will be the area. 

I. What is the area of a parabola whose double ordinate is 24m. 
and altitude 16m. ? 

By formula, A=% (x. 2/)=f( 16X24 )=256m2. 

"1. 24m.=^AT"(in last figure )=the double ordinate, or base 
of the parabola. 



II. 



2. 16m.==^4 =the altitude of the parabola. 



3. /. 384m 2 ==16 X 24=the area of the rectangle circumscribed 

about the parabola. 

4. f of384m 2 =256m 2 =the area of the parabola. 
III. .. The area of the parabola is 256m 2 . 



MENSURATION. 229 

Prob. XXXVIII. To find the area of a parabolic frustum 
having- given the double ordinates of its ends and the dis 
tance between them. 

B* _ 3 

Formula. A=%aXj2 2 __ 2 , in which ais the distance be 

tween the double ordinates, B the greater and b the lesser doubl< 
ordinate. 

Rule. Divide the difference of the cubes of the two ends by the 
difference of their squares and multiply the quotient by \ of the 
attitude. 

I. What is the area of a parabolic frustum whose greater base 
is 10 feet, lesser base 6 feet, and the altitude 4 feet? 

By formula, A= 

(1. 10 ft.=the greater base, 

2. 6 ft.=the lesser base, and 

3. 4 ft.=the altitude. 

4. 784 cu. ft.=10 3 6 3 =the difference of the cubes of the 
j i two bases. 

' 5. 64 sq. ft.=10 2 6 2 =the difference of the squares of the 

two bases. 
6. 12 ft.=784-i-64=the quotient of the difference of the 

cubes by the difference of the squares. 
17. .', iX(4Xl2i)=32t sq. ft.=the area of the frustum. 
III. .-. The area of the frustum is 32f sq. ft. 

3. HYPERBOLA. 

1. a 2 y 2 6 2 x 2 = a-b' 1 is the equation of the hyperbola referred 
to its axes in terms of its semi-axes. 

32 

2. y*= -- ~(2ax x 2 ) is the equation of a hyperbola referred 

to its transverse axis and a tangent at the left hand vertex. 

3. 0= 4 is the polar equation of the hyperbola. 

1 e cos u 

Having given any three of the four quantites , b, x,y, the oth- 
er may be found by solving the rectangular equation with refer- 
ence to the required quantity. 

Prob. XXXIX. To find the length of any arc of an hyper- 
bola, beginning 1 at the vertex. 



|/<C a 2_l_^2\ r 2 I ^4^ 



. _ 

--*^ 



1 <* 2 * 2 1.1-3 ^ 4 +4^ 2 ^ 2 4 1.1.3.3.5 
+1.2.3 ~b T ~' 1.2.3.4.5 3 8 y ^12.3.4.5.6-7 






230 FINKEL'S SOLUTION BOOK. 

Rule. ! Find the parameter by dividing the square of the 
conjugate diameter by the transverse diameter. 

2. To 19 times the transverse, add 21 times the parameter of 
the axis, and multiply the sum by the quotient of the abscissa di- 
vided by the transverse. 

3. To 9 times the transverse, add 21 times the parameter , and 
multiply the sum by the quotient of the abscissa divided by the 
transverse. 

4. To each of the products thus found, add 15 times the par a- 
qneter, and divide the former by the latter; then this quotient 
being multiplied by the ordinate iv ill give the length, nearly. 

{Bonnycastle 1 s Rule*} 

NOTE. A parameter is a double ordinate passing through the focus. 
I. In the the hyperbola DA C, the transverse diameter GA 
=80, the conjugate 777=60, the ordinate 7?C=10, and the 
abscissa yl^?=2.1637 ; what is the length of the arc DA C? 

By formula, 



1.1.3 u- , ^ ~ 
1.2.3.4.5 b* 
1.13.3.5 



1.2.3.4.5.6-7 




20.658. __^_ 

" FIG. 20, 

1. 45=207^0^=23^ -r-tf=the parameter LK which, in 

the figure, should be drawn to the right of DC, to be 
consistent with the nature of the problem. 

2. 1520=19X80=19 times the transverse diameter. 

3. 945=21X45=21 times the parameter. 

4. 2465=sum of these two products. 

5. .02704=2. 1637-f-80=quotient of abscissa and transverse 

diameter. 

6. 2465 X .02704=66.6536=sum of the products multiplied 

by the said quotient. Also, 



7. 

45.0216. Whence 

8. (15x45+66.6536 )-$-(15x45+45.0216)=741.6536-r- 

720.0216=1.03004. 

9. .-. 1.03004XlO=10.3004=length of the arc A C, nearly. 
III. .-. The length of the arc is 10.3004. 



MENSURATION. 231 

Prob. XIu To find the area of an hyper bola,the transverse 
and conjugate axes and abscissa being: given. 

Formulae. (a)A=2Jydx =&- C x/ (x* a 2 )* dx=.-x' 



; or, (b) A=4xy * 




_f __*!_] J 

.5.7. [a*+x*\ 5.7. 



Rule. 1. To the product of the transverse diameter and 
abscissa, add \ of the square of the abscissa, and multiply the- 
square root of the sum by 21. 

2. Add 4 times the square root of the product of the transverse 
diameter and abscissa^ to the product last found and divide the 
sum by If 5. 

3. Divide Jf. times the product of the conjugate diameter and 
abscissa by the transverse diameter, and this last quotent multi- 
plied by the former will give the area required, nearly. Bonny- 
castlds Rule. 

I. If, in the hyperbola DA C, the transverse axis A G is 30, the 
conjugate diameter HI, 18 and the abscissa AB, 10 ; what is the 
area of the hyperbola DA C ? 



By formula (a),A=x'y'ab log*" ~l= = 25y / 



15X9 lo, e=3 o -1351o =300 



135 log e 3=300 135 Xl.09861228=151.687343,/ being found 
from the equation a 2 / 2 & 2 x /<2 = a*b , in which =15, 3=9 
and *'=15+10=25. 

1. 1. 2lV30XlO+fXl0 2 =2lV300+50(M-7=2lV371.42857 

=21X19.272=404.712, by the first part of the rule. 

2. 2. (4V30xlO+404.712)-i-75=(4xl7.3205+404.712)~ 
H <; 75=6.3199, by the second part of the rule. 

3. 3. .'. 18X 9 1 n X4 X 6.3199=151.6776, by the third part 

oU 
of the rule, =the area of the hyperbola, nearly. 

III. .-. 151.6776=the area oftbs " T -.-"^ 



-232 . FINKEL'S SOLUTION BOOK. 

Prob. XLiI, To find the area of a zone of an hyperbola. 

b r** 



Formula. A=2- ?***-** 

V 




'i 

in which (x 2 ,y. 2 ), and (#j, y l ) are the co-ordinates of the points 
<C and L, respectively. 

I. What is the area of a zone of an hyperbola whose transverse 
diameter is 2=10 feet, conjugate diameter 2=6 feet, the lesser 
'double ordinate of the zone being 8 feet and the greater 12 feet? 

By formula, A= * 2 JK 2 x \y\ a ^& 



But from the equation, when jy==y 2 =6, x==x 2 =W\ / Q and when 
y=y 1 =4:,x=x 1 =13%. Substituting these values of x 2 andjy 2 



we have ^=506- 



= 50V6 66 1 301og <? [f(V6 + l)] sq. ft. 

Prob. XLII. To find the area of a sector of an hyperbola, 
K.ALO. 

Formula. A=at> lo 

Rule. Find the area of the segment A KL by Prob. XL t> and 
subtract it from the area oj the triangle KOL. 

I. What is the area of the sector OAL (Fig. 20) if OA=a=5, 
Ol=b =3, and L F=4: ? 



By formula, A=%ab lo 

f 20)] But when y==4, ^=134. Hence, 



MENSURATION. 

IX. HIGHER PLANE CURVES. 



233- 



1. Higher Plane Curves are loci whose equations are 
above the second degree, or which involve transcendental func- 
tions, /. ., a function whose degree is infinite. 

I. THE CISSOID OF DIOCLES. 

1. The Cissoid of Diodes is the curve generated by the 
vertex of a parabola rolling on an equal parabola. 

2. If pairs of equal ordinates be drawn to the diameter of a. 
circle, and through one extremity of this diameter and the point 
in the circumference through which one of the ordinates is let 
fall, a line be drawn, the locus of the intersection of this line and 
the equal ordinate, or that ordinate produced is the Cissoid 
of Diodes. 

x 8 

3. y 2 =- is the equation of the cissoid referred to rectan- 

2a x 

gular axes. 

p=2a sin# tan# is the polar equation of the curve. 

Prob. XLIII. To find the length of an arc GAP of the 
cissoid. 



Formula. s= OAP= 




r 2(3+2) -I) 
LVs V20 *V80 3* J ' 



I. What is the length of the arc 
OAN, in which case 



By formula, s=a J 




Fig. 21. 



234 FINKEL'S SOLUTION BOOK. 

Prpb. XLIV. To find the area included between the curve 
and its asymptote, BM. 



Formula 



_ 

/** /*2 i #3 p 

. A=2 I ydx=2 I -J- - dx=\ J 

Jo Jo ^&a x L. 



- =3?r 2 , *.*., 3 times the are 




of the circle, OEB. 

Note. The name Cissoid is from the Greek xtffffoeidlq, like ivy, fron 
x t ff a 6 c, ivy, eldo<; form. The curve was invented by the Greek geometeJ 
Diodes, A D. 500, for the purpose of solving two celebrated problems ol 
the higher geometry; viz., to trisect a plane angle, and to construct twc 
geometrical means between two given straight lines. The construction ol 
two geometrical means between two given straight lines is effected by the 
cissoid. Thus in the figure of the cissoid, ED and OG are the two 
geometrical means between the straight lines OD and JPG: that is, 
OD:ED\'.OG\PG. The trisection of a plane angle is effected by the con- 
choid. The duplication of the cube, i. <?., to find the edge of a cube whose 
volume shall be twice that of a cube whose volume is given, may be effected 
by the cissoid. Thus, on KC lay off CH=2BC, and draw BH. Let fall 
from the point F, where B H cuts the curve, the perpendicular FR. Then 
RF=&BR. Now a cube described on RFis twice one described on OR', 

OR* OR* 
for, since FR=y,OR=x, and l?R=2ax, we have RF* ^^ ^ D , or 

J3K 3-F-K 

\RF*=OR*. .'. FR*=20R* O^ E. D. 

2. THE CONCHOID OF NICOMEDES. 

1. The Conchoid is the locus formed by measuring, on a 
line which revolves about a fixed point witbout a given fixed 
line, a constant length in either direction from the point where it 
intersects the given fixed line. 

2. x z y 2 =(&-}-y) 2 (a 2 y 8 ) is the equation of the conchoid re- 
ferred to rectangular axes. 




FIG. 22. 

3. p=b sec 8a\s the polar equation referred to polar co-ordi- 
nates. In this equation, is the angle PO makes with A O. 

Prob. XLV. To find the length of an arc of the conchoid. 



Formula. s= l+ j <#=jVl+tan 6 sec<#. 



MENSURATION. 



235 



Prob. XLVI. To find the whole area of the ciiichoid be- 
tween two radiants each making an angle 8 with OA. 

Formula. A=2 %r*dO=b* J*(sec 8a)*d6= 

tan 0-{-2a 2 8, according as a is or 



6 2 tan 

is not greater than b. 



or 



1. The area above the directrix m m' and 
the same radiants =203 log tan ( -r -|- - j-|-<z 2 #. 

The area of the loop which exists when a is 



2. 



. o z i 
cos- --2^ log 



b is a 2 



NOTE. The name conchoid is from the Greek, xor%oet8s t from 
jloyXr), shell, and sT.8oq } form, and signifies shell-form. It was invented 
by the Greek geometer Nicomedes, about A. D. 100 for the purpose of tri- 
secting any plane angle. The trisection of an angle may be accomplished 
by this curve as follows: Let A OH be any angle to be trisected. From any 
point, G, in one side let fall a perpendicular, GF, upon the other. Take 
AF=2 GO, and with O as the fixed point, m m' as the fixed line and /'Pas the 
revolving line of which Piy=.a is constant, construct the arc of the con- 
choid, PAH. Erect BG perpendicular to mm' and draw BO. Then is 
BOA one third of HO A. For bisect B Kat /, and draw GI. Also draw IL 
parallel to GK. Since BIIK, BL=LG and GI=BIIK^=GO. By 
reason of the isosceles triangle BIG, we have the angle GIO=!2 GBO~=. 
2LBQA. But l_GIO=LIQG. : 2IOA^/_2QG, or IOA %/_HOA. 

C^, E. F. 

3. THE OVAL OF CASSINI OR CASSINIAN. 

1. The Oval of Cassini is the locus of the vertex of the 
triangle whose base is 2a and the product of the other sides =m*. 

2. 



=m 4 is the rectangular equation of the curve, in which 



3. r 4 2a 2 r 2 co$26-\-a 2 m*=Q is the polar equation of the 
curve. 

Discussion. If a be ^> m^ 
there are two ovals, as shown 
in the figure. In that case, the 
last equation shows that if 
OPP' meets the curve in P 
and P', we have OP.OP'= 
V<z 4 m^\ and therefore the 
curve is its own inverse with 
respect to a circle of radius= 
VV_a4 fflHUBSM 

FIG. 23. 




FINKEL'S SOLUTION BOOK. 



4. LEMNISCATE OF BERNOUILLI. 

1. This curve is what a cassinian becomes when ma. 
above equations then reduce to 

2. ( 

3. r 

Prob. XLVII. To find the 
length of the arc of the Lemnis- 
cate. 



The- 



Formula. s== 




FIG. 24. 



C nT^YT^V/gL-/^ 2 ^ i a a * dr 

= I -J^ 2 H si 1 - 4 }dd= I dd= I -- =-.= 
J ^ ' r^V a >/ J r J Vtf 4 r 4 ) 



*f So 4+toAjn + &c -} When ^ =fl 

_|_^.|.|.. T i r -)-&c"|= arc BPA. .'. The entire length of the curve i 



Prob. XL. VIII. To find the area of the lemniscate. 
Formula. A=4 hr*ct0=4a* A 7r cos26 l d0 






5. THE VERSIERA OR WITCH OF AGNESI. 



1. The Versiera is the locus of the extremity of an ordi- 
nate to a circle, produced until the produced ordinate is to the 
ordinate itself, as the diameter of a circle is to one of the seg- 
ments into which the ordinate divides the diameter, these seg- 
ments being all taken on the same side. 

2. Let P be any point of the curve, PD=y, the ordinate of 
the point P and 

OD=X) the abscis- 
sa. Then, by defini- 
tion, EP \EF\\ 
AO\EO, or x: 
EF : :2a \y. But 




2a :y. Whence x*y= FIG. 25. 

y) is the equation referred to rectangular co-ordinates. 



MENSURATION. 237 

3. r ( r * r * sin 2 6-\-4a 2 ) sin6=8a* is the polar equation of 
the curve. 

Prob. XLIX. To find the length of an arc of the Versiera. 

^ 

This can be integrated by series and the result obtained ap- 
proximately. 

Prob. L. To find the area between the cttrve and its 
asymptote. 



Formula. 



. A=2 A/^=2X8 3 /*** , f * a = 
J* Jo x*--4a 2 



20 J 
Rule. Multiply the area of the given circle by 4. 

NOTE. This curve was invented by an Italian lady, Dona Maria Agnesi, 
1748. 

6. THE LIMACON. 

1. The Limacon is the locus of a point P on the radius 

vector OP) of a circle OFB from a fixed 
point, <9,on the circle and at a constant 
distance from either side of the circle. 

2. ( x *+y*ax)* = l>*(x*+y*) is 
the rectangular equation of the curve. 

3. r=acos9b\* the polar equa- 
tion. In these equations, a= OA and 



Prob. LI. To find the length of 
an arc of the Limacon. 

Formula. s= 




FIG. 26. 
ibcos fy)c 



J\ j (a+&) 2 cos^ e - f A-(d bysin*^ \ d8. .-. The rectification 



of the Limacon depends on that of an ellipse whose semi-axes are 
(a-\-b) and (a b.) 

When a=t>, the curve is the cardioid, the polar equation of 



which is r=a(l+cos 6 ), and 



=J ^ r2 +(jj$ 



238 



FINKEL'S SOLUTION BOOK. 



OdO 



C^Tt 

2a I cos 4- 6 d&=&a=the entire length of the cardioid. 

J 7t 

Prob. LJI. To find the areaof the Liimacon. 

* 



Formula. A= 



=\ (a cos 0+&)*d0= 



. When a=b, the curve becomes a cardioid, and A= 
%7ra 2 . When cT>b, the curve has two loops and is that in the figure. 
r=acosO-\-b is the polar equation of the outer loop, and r=a 
CO s b is the polar equation of the inner loop. The area of the 

_i ft 
inner loop is A=J%r*d(t=% C <l (acosQbyde= 



NOTE. This curve was invented bj Blaise Pascal in 1643. When 
2, the curve is called the Trisectrix. 

7. THE QUADRATRIX. 

1. The Quadratrix is the locus of the intersection, P,of 
the radius, OD, and the ordinate 

QN, when these move uniformally, 
so that ON:OA:\l_BOD:%n. 

fa x n\ 

2. y=xtanl - .- 1 is the 

rectangular equation of the curve, 
in which aOA, x=ON, and 



3. The curve effects the quad- 
rature of the circle, for OCiOBr. 
rc ADB. 




Prob. LIII. To find the area 
enclosed above the x axis. 

Formula. A=Jydx= FIQ 2J 

/v tan I .- }dx=4:a 2 7r~ 1 log 2. 
\^ a % J 
** * ** ^ 

NOTE. This curve was invented by Dinostratus, in 370 B. C. 

8. THE CATENARY. 

1. The Catenary'^ the line which a perfectly flexible chain 
assumes when its ends are fastened at two points as B and C in 
the figure. 



MENSURATION. 239 

a> X 

2. y=^a(ea-\-e~a) is the rectangular equation of the curve, in 
which a= OA. A is the origin of co-ordinates. BAC is the 
catenary. M*APM\& the 
evolute of the catenary 
and is called the Trc- 
trix. To find the equa- 
tion of the curve, let A be 
the origin of co-ordinates. 
Let s denote the length 
of any arc AE ; then, if p 
be the weight of a unit 
of length of the chain, 
the verticle tension at E, 

is sp. Let the horizontal FIG. 28. 

tension at E, be ap, the weight of a units of length of the chain. 
Let EG be a tangent at E, then, if EG represents the tension of 
the chain at B, EF and GF will represent respectively its hori- 
zontal and its vertical tension at B. 




___ . 
' dx~EF~ap~a '' a 



/ ds 

' x=a J v* 



(s-\-* a 2 +s 2 )-\-c. Since x=o, when s=o, c= 




x=a log-+>l4-- From this equation, we find s= 



af to - x *\ 

2\f"~^ e J which is the length of the curve measured from A. 

dy s dy s f * X 
But-/=-. /. -f =- = 41 Ji e -a 1. 

dx a dx a ^\ e J 



Prob. I<IV. To find the area of the Catenary. 

C ae an 

Formula. A=Jydx=J%a \ ^+ e ^ 

( 25X : - 

e a-\-i~a J=a\y 2 a*. This is the area included between the 

axis of #, the curve and the two 



NOTE. The form of equilibrium of a flexible chain was first investigated 
by Galileo, who pronounced the curve to be a parabola. His error was de- 
tected experimentally, in 1669, by Joachim Jungius, a German geometer; 
but the true form of the Catenary was obtained by James Bernouilli, in 1691. 



240 FINKEL'S SOLUTION BOOK. 

9. THE TRACTRIX. 

1. The Tractrix is the involute of the Catenary. 

( . ) . 

2. x=a log 0-j-V(08_- y 2 ) a logy V( 2 y 2 , is the rect- 
angular equation of the curve. 

I*rob. LV. To find the length of an arc of the tractrix. 

Formula. s a log ( - Y 

Prob. I/VT. To find the area included by the four branches* 

/ 

Formula. A=Jydx= 4J o \a 2 y 2 dy=7ta 2 . 

1C. THE SYNTRACTRIX. 

1. The Syntractrix is the locus of a point, Q, on the tan- 
gent, PT, of the Tractrix. 

2. x= a log | c+</(c z y 2 ) I a log/ V(c 2 j/ 2 ) istherect- 

angular equation of the Syntractrix, in which c is QT 1 , a con- 
stant length. 

11. ROULETTES. 

1. A. Houlette is the tocus of a point rigidly connected 
with a curve which rolls upon a fixed right line or curve. 

(a) CYCLOIDS. 

1. The Cycloid is the roulette generated by a point in the 
circumference of a circle which rolls upon a right line. 

2. A Prolate Cycloid is the roulette generated by a point 
without the circumference of a circle which rolls upon a right 
line. 

3. A Curtate Cycloid is the roulette generated by a point 
within the circumference of a circle which rolls upon a right line. 

4. x=versin~ l y *J%ry -y 2 is the rectangular equation of the 
cycloid referred to its base and a perpendicular at the left hand 
vertex. To produce this equation, let AN=x and PN=y, P be- 
ing any point of the curve. Let OC 

r=the radius of the generatrix OPL. 
Now AN=A ONO. But by con- 
struction A O=arc P O=versin~ v 
or versin~*y to a radius r. 





YI fy 

Or, we may have xa( 6 sinO) , and FIG. 29. 

y=a(l=cos 6) in which 6 is the angle, PCO, through which the 
generatrix has rolled. 



MENSURATION. 241 



For x=AO NO. But AO=aLP CO=aO, and 
= PC sin/_PCF=a sinQ. .-. x=aOa sin6=a(6sin8) , y= 
OCCF=aCF. But CF=PC cos LPCF^a cos 
a a cosO=a(\ cosB). 

Prob. LVII. To find the length of an arc of the cycloid. 



*&r J (2r y)~ k dy= 2V2r(2r y)*+c. Reckoning the arc from 
the origin, c4r; and the corrected integral is s= 2(2r) 

(2r _y)*4-4r. Whenj/=2r, s=4r. .-. The whole length of the 
cycloid is 8^=4Z?, i e.^ the length of the cycloid is 4 times the 
diameter of the generating circle. 



. (1) Multiply the corresponding chord of the genera* 
trix, by 2. To find the length of the cycloid : 

(#) Multiply the diameter of the generating circle by 4- 

I. Through what distance will a rivet in the tire of a 3-ft. 
buggy wheel pass in three revolutions of the wheel? 

By formula, 5t=3(S>)=24xlift.=36ft. 

1. 3ft.=the diameter of the wheel. Then 

2. 12ft.=4x3ft.=di*tance through which it moves in 1 



II. 



revolution. 



3. .-. 36ft.=3Xl2f't.=distance through which it moves in 

3 revolutions. 
III. .-. It will move through a distance of 36 ft. 

Prob. ITVIII. To find the area of a cycloid. 

Formula. A=2 fyctx=2 f ' y *-^ = 
J 'o V2ry y* 



Rule. Multiply the area of the generating circle by 3. 

I. What is the area of a cycloid generated by a circle whose 
radius is 2ft. ? 

By formula,^4=37rr 2 =37r2 2 ==127T=37.6992 sq. ft. 
( 1. 2ft.=the radius of the generating circle. 

II. < 2. 7r2 2 :=12.5664 sq. ft=the area of the generating circle. 
( 3. 37r2 2 =37.6992 sq. ft.=the area of the cycloid, 

III. .-. The area of the cycloid is 37.6992 sq. ft. 

Prob. L.IX. Awheel whose radius is r rolls along- a hori- 
zontal line with a velocity v'; required the velocity of any 
point, P, in its circumference; also the velocity of P horizon- 
tally and vertically. 



242 FINKEL'S SOLUTION BOOK. 

Since a point in the circumference of a wheel describes, in 
space, a cycloid, let P, Fig. 29, be the point, referred to the 
axes A A' and a perpendicular at A. Let (x,y) be the coordi- 
nates of the point; then will the horizontal and vertical 
velocities of P be the rates of change of x and y respectively. 

\t 

O being the point of contact, A Or versin~ 1 . Since the cen- 

ter C, is vertically over 0, its velocity is equal to the rate of in- 
crease of A 0. In an element of time, dt, the center C will move 

(y^ rdy 
r versiri~* J - I /- - ==. .'. Its velocity v' = 
rj \2ry y 2 

the distance it moves divided by the time it moves, or v'= 



dy r dy dy *lry y 2 , ., 

J -r-dt=-p= - . -. .'. - = J J T/= the velocity 
\2ry _ -y 2 dt at r 

vertically. . . . (1), 



From the equation of the cycloid, x=rversin~ l -- *J%ry -j 2 , we 

v 
have dx=.- J dy. Now dx-^-dt=\hQ velocity of the point 

* 



horizontallv. But dx-^dt, or =-.=- ==. -~. Substituting the 

df \/2ryy 2 dt 

value of ~, we have =.-v' ..... (2). An element of the 
dt dt r 

curve APBA' is ds and this is the distance the point travels 

in an element of time, dt. .*. = the velocity of the point, P. 

dt 



But ds=Vd*dx*=* + v 'dt= v'dt, since, 



from (1), dy= and, from (2), fr. .'. By 



dividing by dt, we have --=v^-v'== the velocity of the 
point, P ...... (3). From (1), (2), and (3), we have, 



MENSURATION. 243 

Hence, when a point of the circumference is in contact with 
the line, its velocity is 0\ when it is in the same horizontal plane 
as the center, its velocity horizontally and verically is the same 
as the velocity of the center, and when it is at the highest point, 
its motion is entirely horizontal, and its velocity is twice that of 

ds J2y /y/2ry 

the center. Since -=- = NL i/ = - -v' ' we have by proportion, 
dt N r r 

- : v' : : V^\r. But V=PF*FO* =PO. 



.-. The velocity of P is to that of C as the chord PO is to the 
radius CO; that is, P and C are momentarily moving about O 
with equal angular velocity. 

(b) THE PROLATE AND CURTATE CYCLOID. 

1. x=a(U m sin#), y=a(l mcosO) are the equations in 
every case. 

2. The cycloid is prolate when m is >1 as AfP'STA', Fig. 
30, and curtate when m is <C1, as PB. These equations are found 
thus: Let CP=ma, and tOCP=0. Thenx=AN=AOOW. 
But AO=i\rc subtended by /.OCP=a6, and ON=PCx*fa 
LNPC=ma*\K LNPC(=PCL=7t ( J)=ma sin (n8)= 
ma sin#. .-. x=atfma sin6=a(6 m sin#). y=PN=OC-\~ 
PC C05/.NPC (=PCL=7rV)=a+macos(7r0)=a 
ma cos^ =a( 1 m cos Q ). The same reasoning applies when we 
assume the point to be P'. 

NOTE. These curves are also called Trochoids. 
Prob. L.X. To find the length of a Trochoid. 
Formula. s= 



Since x=a(6 m sin#), dx=a(l mcosP)dtf] and since y= 
a(\m cos6> ), dy=am sin^ft /. s= 




I. If a fly is on the spoke of a carnage wheel 5 feet in diame- 
ter, 6 inches up from the ground, through what distance will the 



244 



FINKEL'S SOLUTION BOOK. 



fly move while the wheel makes one revolution on a level plane? 

Let Cbe the center of 
the wheel, in the figure, 
and P the position of the 
fly at any time. Let O C= 
the radius of the carriage 
wheel =0=2$ ft, PC= 
2ft, and the angle OCP 
=0. Let (x,y) be the co- 
ordinates of the point P. 
Let F, a point at the inter- FI& 30. 

section of the curve and AI be the position of the fly when the 
motion of the wheel commenced. Then since xa(V msinff) 
and y=a(\ mcosO), we have dx=a(l mcoB0)du t and dy= 




a m s 



N | 



dO^ 



cos 



(1+ 



snce P ==ma= 



, in which <p=\Q. But 7^=2 ft., and 

/ITT 

ft. , m== 2-7-2^=4. ' ^=4 X 2^ I 

Jo 



18.84 ft. 



II. .-. The fly will move 18.84 ft. 



Prob. !LXI. To find the area contained between the tro- 
choid and its axis. 



Formula. A=<tx= 



1m cos 6) ( 1 



m cos 



cos 



in6cos6)^\ 7r =2a 2 (7r-\-%m 2 7r). When w=l, the curve is 

So 

the ccloid and the area=37r 2 as it should be. 



* When <pis replaced by (\n-\-<p), this is an elliptic integral of 
the second kind and ma be written 



MENSURATION. 



(c) EPITROCHOID AND HYPOTROCHOID. 

1. An Epitrochoid is the roulette formed by a circle roll- 
ing upon the convex circumference of a fixed circle, and carrying 
a generating point either within or without the rolling circle. 

2. An Hypotrochoid is the roulette formed by a circle 
rolling upon the concave circumference of a fixed circle, and 
carrying a generating point either within or without the rolling 
circle. 

3. *=(#-|-)cos 9 ml>cos-~-0, y=(a-\-b) sin 6 mb sin -- 6 

are the equations of the epitrochoids. 

In the figure, let C be the center of the fixed circle and O the 
center of the rolling circle. Let FP'Q be a portion of the curve 
generated by the point P' situated within the rolling circle, and 
let CG=x and P' G=y be the co-ordinates of the point, P / '. 

Let A be the position of P when the rolling commences, and 
q*=/_POC through which it rolled. Draw OK perpendicular 
to C G and P'l perpendicular to OK; draw Z>P and DP'. Let 
OP'=mOP=mb and the angle A CD=6. Then x=CG=CA 
. But Cfe= 0C cos #==(04-3) cos and 



\7t-(<p+)\=-mb 
But arc 



AD=arcPD. .-. 
bcp. Whence (p=~6. 



b 

and x=(a-\-l>) cos 

/>_!_/> 
mb cos 



b 
IK=OKOL But 




, and OI=OP / sin/ OP'I=mb sin] n (<p+B) \ 

n 6 



If f0=l, the point jP x will be on the circumference of the roll- 
ing circle and will describe the curve APN which is called the 



246 FINKEL'S SOLUTION BOOK. 

Epicycloid The equations for the Epicycloid are #=(tf 

cos-^t-#, and y==(a4-b) sin d sin j- 6. The equations for the 
o o 

Hypotrochoid may be obtained by changing the signs of b and mb, 
in the equations for the Epitrochoid. .'. x(a )cos B -\-rnb cos 

6, andy=(a )sin mbsin 7 6 are the equations for the 

Hypotrochoid. If m=l, the generating point is in the circumfer- 
ence of the rolling circle and the curve generated will be a 



a 



cycloid. .'. x=(a b) cos# -\-b cos y '/, and y=(a )sin 6 
b sin 9 are the- equations of the Hypocycloid . 

Prob. LiXII. To find the length of the arc of an epitro- 
choid. 



For mula. s=J*ldx*-\-dy*= | V( a +b) sin 

s Q m(a+b) cos 



a+6) f J 



cos 



<?0=(a+6) i(l+ m * 2mco d)dB. This 



may be expressed as an elliptic integral, E(k, cp), of the second 
kind, by substituting (TT -| cp) for 0, and then reducing. 

2. By making /=1, we have s=(a-{-l>)\ / 2 f'fl^L 
co&y ft) dQ, the length of the arc of an hypocycloid. 

3. By changing sign of b, the above formula reduces to y=s 
(a b) jY(14-^ 2 +2#2cos^ (9 )dO, which is the length of the arc 
of an hypotrochoid. 

4. By making m=l, in the last formula, we have s= 

(a <5)V2 f(l+cos - &)%dd, which is the length of the arc of - 
hypocycloid. 

I. A circle 2 ft. in diameter rolls upon the convex circmii' 
ference of a circle whose diameter is 6 feet. What is the length 



MENSURATION. 24T 

of the curve described by a point 4 inches from the center of the 
rolling circle, the rolling circle having made a complete revolu- 
tion about the fixed circle ? 

In Fig. 31, let Obe the center of the rolling circle; C the center 
of the fixed circle ; CZ>=3 ft.=a, the radius of fixed circle; OD= 
lft.=, the radius of the rolling circle; OP=4 inches=ij- of 12 
inches 772^ the distance of the point from the center; and P the 
position of the point at any time after the rolling begins. Let 6= 
the angle A CD and <p=the angle POD through which the roll- 
ing circle has rolled. Then we have, as previously shown, the 
equations of the locus P, 

x=(a+b)cos 6ml> cos(<p-f #)=(-[-) cos 6 mb cos 



o 
mb sin<--0=tf--3sm 6 mb sin 



From these equations, we can find dx and dy, 



_ 

.-. By formula, s=J\dx 2 +(ty2=G(a+&)J V(l+/w 2 
2 cos ? 0)<* 0=24 /^VU+Ci) 2 I cos30)*fl=8 T 

O J o Jo 

6 cos 30X<9. Let 3 0=2$. Then ^=8 /^(lO 6 cos 3 

/0 

s^)^, =2Hf^ [l i-l cos*0 i.i-( 

COS ^-fi t- f (I) 4 COS^- 



(|) 4 &c. =10|7TX .773=26.9 ft, nearly. 



Remark. When the point is on the circumference of the roll 
ing wheel, the length of the curve generated by the point is 



If we let the conditions of the above problem remain the same,. 



248 FINKEL'S SOLUTION BOOK. 

only changing the generating point to the circumference, we have 
for the length of the curve, *=6V2(3+1) /^Vfl cos3ff)d6= 

48 / 2 V(l i cos 2 (p )</<>, where <p=f #. Expanding this by the 
Binomial Theorem and integrating each term separately, 
24* l- 



I. A circle whose radius is 1 ft. is rolled on the concave cir- 
cumference of a circle whose radius is 4 ft. What is the length 
of the curve generated by a point in the circumference of the 
rolling circle, the rolling circle having returned to the point of 
starting? 

-.(a )cos 8-\-b cos - 0, 



^jr _ ^ 

y=(a b) sin 6 b sin 0, are the equations of the curve 

which is a hypocycloid. In these equations #=4 and 3=1. 
.-. x=3 cos 0+cos30=4cos 8 0, and 
sin sin 3 0=4 sin 3 0. Whence, 

nd.infl= 



cos 2 0+sin 2 ft=-|_ . But cos 8 0+sm 2 0=1. 



=43, which is the rectangular equation of the curve. 



By formula,5=(^ \-dy* )=4 



=-6x4=24 tt 



X. PLANK SPIRALS. 

1. ^4. Plane Spiral is the locus of a point revolving about 
a fixed point and continually receding from it in such a manner 
that the radius vector is a function of the variable angle. Such 
a curve may cut a right line in an infinite number of points. 
This would render its rectilinear equation of an infinite degree. 
Hence, these loci are transcendental. 

2. The Measuring Circle is the circle whose radius is 
the radius vector of the spiral, at the end of one revolution of the 
generating point in the positive direction. 



MENSURATION. 



249' 



3. A Spire is the portion of the spiral generated by any 
one revolution of the generating point. 

1. THE SPIRAL OF ARCHIMEDES. 

1. The Spiral of Archimedes is the locus of a point 
revolving about and receding from a fixed point so that the ratio* 
of the radius vector to the angle through which it has moved 
from the polar axis, is constant. 

2. r=aO is the polar equation of this curve. 

Prob. L.XIII. To find 
the length of the spiral 
of Archimedes. 

Formula. s= 



which is the length of the 
curve measured from the 
origin. 




FIG- 32. 



27r-f-V[l+(27r) 2 ] ( is the length 



of the curve made by one revolution of the generating point. 
Prob. LiXIV. To find the area of the spiral of Archimedes. 

Formula. A=$Jr*<t0=$a z J0*<te=la 2 e*=^-, the 
area measured from the origin. 
2. THE RECIPROCAL OR HYPERBOLIC SPIRAL. 

1. The Reciprocal or Hyperbolic Spiral is the locus. 
of a point revolving around and receding from a fixed point so 
that the inverse ratio of the radius vector to the angle through 
which it has moved from the polar axis, is constant. 



2. 



'= is the polar equation of the Hyperbolic Spiral. 



Prob. LXV. To find the length 
of the Hyperbolic Spiral. 

Formula. s= 




FIG. 33. 



250 FINKEL'S SOLUTION BOOK. 

/r'(l+0 2 )i=log j 0+V(~l+# 2 ) I tf-H/1+0 2 , is the length of 
the spiral measured from the origin. 

Prob. LiXVI. To find the area of the Hyperbolic Spiral. 

Formula. A=%Jr* dfi=%a 2 J--=^, the area 

measured from the origin. This result must be made positive 
since the radius vector revolves in the negative direction. 

3. THE LITUUS. 

1. The LitUUS is the locus of a point revolving around 
and receding from a fixed point so that the inverse ratio of the 
radius vector to the square root of the angle through which it has 
anoved, is constant. 

a 

2. r =\7 i s tne equation of the 




Lituus. 

Prob. LXVII. To find the length 
of the Lituus. 

FIG. 34 

Formula. s= * 



Prob. LXVIII. To find the area of the Lituus. 

Formula.~A=^Jr^0=^C~=^ log 6. 

4. THE LOGARITHMIC SPIRAL. 

1. The Logarithmic Spiral is the locus genera ted by a 
point revolving around and receding from a fixed point in such 
a manner that the radius vector increases in a geometrical ratio, 
while the variable angle increases in an arithmetrical ratio. 

2. r=a e is the polar equation of the Logarithmic Spiral. If 
is the base of a system of logarithms, this equation becomes 

r. 



Prob. LXIX. To find the length of the Logarithmic 
Spiral. . 




MENSURATION. 251 

r, where m is the modulas of the system 
of logarithms. 

Prob. LXX. To find the area of the 
Logarithmic Spiral. 

Formula. A=$Jr*d8=-~ I rdr= 

\mr*. Since m=l, in the Naparian System 
of Logarithms, A=%r 2 , i. e., the area is of 
the square of the radius vector. 

FIG, 35 

XI. MENSURATION OF SOLIDS. 

Prob. LXXI. To find the solidity of a cube, the length oi 
its edge being given. 

Formula. F=(edge) X (edge )X( edge )=( edge) 8 . 

Rule. Multiply the edge of the cube by itself, and thai 
product again by the edge. 

I. What is the volume of a cube whose edge is 5 feet? 
By formula, F=(edge) 3 =(5) 3 =125 cu. ft. 

jj ( 1. 5 ft.=the edge of the cube. 

I 2. 5X5X5=125 cu. ft.=the volume of the cube. 

III. .-. The volume of the cube is 125 cu. ft. 

Remark. Some teachers of mathematics prefer to express th<s 
volume by saying 5x5X5x1 cu. ft. =125X1 cu. ft=125 cu. ft. 



Prob. ixll. To find the volume of a cube, having given 
its diagonal. 

( 

Rule. Divide the diagonal by the square root of 3, and the 
cube of the quotient will be the volume of the cube. 

What is the volume of a cube whose diagonal is 51.9615 inches? 



FINKEL'S SOLUTION BOOK. 



r i v /^Y /^l.eeiSY /^51. 
By formula, V) ={-- ) ^j 



27,000 cu. in. 

II. 51.9615 in.=the diagonal. 
2. 30 in.=51.9615 in.-M/3==51.9615 in.-M.73205=the edge 
of the cube. 
^3. .-. 30X30X30=27,000 cu. in.=the volume of the cube. 

III. /. The volume of the cube whose diagonal is 51.9615 in., 
is 27,000 cu. in. 

Prob. LXXII1. To find the volume of a cube whose surface 
is given. 

Formula 



. F=y^- j . 



Rule. Divide the surface of the cube by 6 and extract the 
squae root of the quotient. This will give the edge of the cube* 
The cube of the edge will be the volume of the cube. 

I. What is the volume of a cube whose surface is 294 square 
feet? 



By formula, V=\ ^J^- ) = 

243 cu. in. 

1. 294 sq. ft.=the surface of the cube 

2. 49 sq. ft.=294 sq. ft.-r 6=area of one fcioe of the cube, 
" ] 3. V49= 7 ft.=length of the edge of the cube. 

t. .'. 7x7x7=343cu. ft.=volume of cube- 
Ill. .'. 343 cu. ft. is the volume of a cube whose surface is 
294 sq. ft. 

Prob. LXXIV. To find the solidity of a parallelopipedon. 

Formula. F=/X^XA where /^length, =breadth, 
and /=thickness. 

Rule. Multiply the length, breadth and thickness together. 

I. What is the volume of a parallelopipedon whose length is 
24 feet, breadth 8 feet, and thickness 5 feet? 
By formula, V=lx&X =24x8x5=9 60 cu. ft. 

(1. 24 ft=the length. 
2. 8 ft.=the breadth, and 
3. 5 ft.=the thickness. 
4. .-. 24X8X5=960 cu. ft.=the volume. 
III. /. 960 cu. ft.=the length of the parallelopipedon. 



MENSURATION. 253 

Prob. LXXV. To find the dimentions of aparallelopite 
don, having- given the ratio of its dimensions and the volume. 



Formula. 1=[ V- 

-, and t=$/[ V-i-(mXnXp)]p, where m,n, and p 
are the ratios of the length, breadth, and thickness respectively. 

Rule. Divide the volume of the parallelopipedon by the pro- 
duct of the ratios of the dimensions, and extract the the cube root 
of the quotient. This gives th^ G. C. D. of the three dimensions. 
Multiply the ratios of the dimensions by the G. C. D. , and the 
results 'will be the dimensions respetively. 

I. What are the dimensions of a parallelopipedon whose 
length, breadth and thickness are in the ratios of 5, 4 and 3; 
and whose volume is 12960 cu. ft. ? 

By formula, /=yfl2960-H(5x4x3)]5=30 ft; =^[12960-- 
(5x4x3)]4=24ft. ; and /=y[12960-^(5x4x3)]3=18 ft. 

1. 5=the quotient obtained by dividing the length by the 

G. C. D. of the three dimensions. 

2. 4=the quotient obtained by dividing the breadth by the 

G. C. D. of the three dimensions. 

3. 3=the quotient obtained by dividing the thickness by 

the G. C. D. of the three dimensions. 

4. .-. 5xG. C. D.=the length, 

5. 4XG. C. D.=the breadth, and 
II A 6. 3XG- C. D.=the thickness. 

7. .-. (5XG. C. D.)X(4XG.C.D.)X(3XG.C.D.)=60X 

(G. C. D.) 3 the volume of the parallelopipedon. 

8. .-.60(G. C. D.) 3 =12960cu. ft. 

9. (G. C. D.) 3 =12960-7-60=216. 

10. /. G. C. D. f 216=6. 

11. .'. 5X(G. C D.)=5X 6=30 ft.=the length, 

12. 4X(G. C. D.)=4X6=24 ft.=the breadth, and 

13. 3X(G. C. D.)=3X6=18 ft.=the thickness. 

III. .-. 30 ft., 24 ft., and 18 ft. are the dimensions of the par- 
allelopipedon. 

Prob. LXXVI. To find the convex surface of a prism. 

Formula. S=pXa, in which/ is the perimeter of the 
base and a the altitude. 



Rule. Multiply the perimeter of the base by the altitude. 



254 FINKEL'S SOLUTION BOOK. 

I. What is the convex surface of the prism ABC Z?, if the 
altitude AE is 12 feet, AB, 6 feet, A C, 5 feet, and BC, 4 feet.? 

By formula, 5=X/=12X(9+5+4)=180 sq. ft. 

( 1. 12 ft =the altitude of the prism. 

II. < 2. 6 ft. +5 ft.+4 ft.=15 ft.=the perimeter of the base. 

( 3. .'. 12x15=180 sq. ft.=the convex surface of the prism. 

III. .-. The convex surface of the pris'm is 180 sq. ft. 

Remark. If the entire surface is required; to the convex sur- 
face, add the area of the two bases. 

Formula. T=S-\- f lA< where 2A is the area of the 
base, S the convex surface, and 7* the total surface. 

Prob. LiXXVII. To find the volume of a prism. 

Formula. F #XA where A is the area of the base, 
<z, the altitude. 

Rule. Multiply the area of the base by the altitude. 

I. What is the volume of the triangular prism ABC Z>, 
whose length AE is 8 feet, and either of the equal sides AB> 
BC, or AC, 2 feet? 

By formula, F===X^=^SX{(2i) t iV3]==i2^5=21.6606 cu. ft- 

1. 8 ft.=the altitude AE. 

2. 21 t"t.=the length of one of the equal 

sides of the base, as AB. 
II.<{3. (2|) 2 iV8=the area of the base ABC, 

by Prob. XL 

4. .-. 8X(2J) 2 iV3=12\ / 3=21.6506cu. ft. 
=the volume of the prism. 

III. .'. -21.6506 cu, ft.=the volume of the 
prism. 

FIG. 36. 

1. THE CYLINDER. 
Prob. LXXVIII. To find the convex surface of a cylinder. 

Formula. S=^a X C, in which a is the altitude and C the 
circumference of the base. 

Rule. Multiply the circumference of the base by the altitude. 

I. What is the convex surface of the right cylinder A GB C, 
whose altitude EF is 20 feet and the diameter of its base AB is 
4 feet? 




MENSURATION. 



255 



By formula, 5=aXC=20x47r=80^=251. 32736 sq. ft 

1. 20 ft.=the altitude EF. 

2. 4 ft.=the diameter AB of the base. 

3. 12.566368 ft.=47T=4x3.141592=the 

circumference of the base. 
4 /. 20x12.566368251.32736 sq. ft.= 
the convex surface of the cylinder. 

The convex surface of the cylinder is 



II. 



III. 



251.32736 sq. ft. 

Remark. If the entire surface is required; to 
the convex surface, add the area of the two 




Formula. T^S+IA^ZnaR+lnR*. FIG. 37. 

Prob. LXXIX. To find the solidity of a cylinder. 
Formula. K ^ 



in which A is the area of the 



base. 



Rule. Multiply the area of the base by the altitude. 

I What is the solidity of the cylinder AGB C, whose alti- 
tude FE is 8 feet and diameter AB of the base 2 feet? 



By formula, F=aX^== 
25.132736 cu. ft. 

II. 8 ft.=the altitude, EF. 
-2. 2 ft.=the diameter, AB, of the base. 
3. 3.141592 sq. ft.=7r/? 2 :=:7rl 2 = area of the base. 
4. .-. 8X3.141592=25.132736 cu. ft. 

III. .'. 25.132736 cu. ft. is the volume of the cylinder. 
2. CYLINDRIC UNGULAS. 

1. A Cylindric Ungula is any portion of a cylinder cut 
off by a plane. 

Prob. LXXX. To find the convex surface of a cylindric 
ung-ula, when the cutting- plane is parallel to the axis of the 
cylinder. 



Formula. S=a J^^ 

arc of the base. 



sn 



Rule. Multiply the arc of the base by the altitude. 

I. What is the surface of the cylindric ungula API Q, 
whose altitude AD is 32 feet and height A T of the arc of the 
base, 2 feet and cord PI of the base 12 feet? 



256 FINKEL'S SOLUTION BOOK. 

By formula, S= a^arc PAI=aZr s\n~ 1 ^=aX^ sin" 1 (^ 

J y \ r 



2AT 



sq. ft, nearly. 



The arc corresponding to the sin 4 is found from a table of 
natural sines and cosines to be (36 



"1 77^1 
of 2 T or ' ' TT. 



a. 2 ft=the height u4 7W the arc 
2. 12 ft.= the length of the chord PL 



3. 12.87 ft.=2V6 2 +2 2 x(l+ 



arc P^7, by Prob. XXV. 

4. .. 32X12.87=411.84 sq. ft. == convex 
surface PAID. 

III. .'. The convex surface of the cylindric 
tmgula PAIQ is 411.84 sq. ft. 

Remark. r is found, by Prob. XX, formula 




FIG. 38. 



Prob. LXXXI. To find the volume of a cylindric ungrila, 
whose cutting plane is parallel to the axis. 



r* ? 



Formula. V=2 C T^ 2 ^* Fdydxdz 2ay( 
Jo J o Jo 



a\r 2 sin" 1 ^ y(r* t y 2 )*\ , in which j|/ is half the chord of the 

base. In this formula f r 2 sin" 1 - j is the area of the sector 

APEIA, 3indy(r 2 y 2 )* is the area of the triangle PEI formed 
by joining the center E with P and /. 

Rule* Multiply the area of the base by the altitude. 

I. What is the volume of the cylindric ungula PIA Z>, if 
PI IB 12 feet, AT2 feet, and altitude AD 40 feet? 



MENSURATION. 257 

By formula, V=aAa \r* sin 1< ^ y(r z y 2 ) | =40|l0 2 sin' l l 









6(10 2 - 6 2 )t=4COOsin ' | 1920=4000^ 1920= 
2574.016- 1920=654.016 cu. ft. 

1. 40 ft.=the altitude AD. 

2. 2 ft.=the height A T of the arc of the base. 

3. 12 ft.=the chord P/of the base. 

II.<J4 16|sq. ft.=^2+l f (2Xl2)=the area of the base, 

by rule, Prob. XXVIII. 

5. .-. 40Xl6i=653 cu. ft.=the volume of the cylindrical 
ungula PIAD. 

III. .'. 653^ cu. ft.=the volume of the cylindrical ungula. 

Remark. A nearer result would have been obtained by finding 
the length of the arc PAI and multiplying it by half the radius. 
This would give the area of the sector IE PA. From the area of 
the sector subtract the area of the triangle PIE formed by join- 
ing P and /with E, and the remainder would be the area of the 
segment PI A. 

Prob. tiXXXH. To find the convex surface of a cylindrio 
nngula, when the plane passes obliquely through the op- 
posite sides of the cylinder. 

Formula. S=^(a-\- a')2?rr, where a and a are the 
least and greatest lengths of the ungula and 2 TIT the circumfer- 
ence of the base of the cylinder. 

Rule. Multiply the circumference of the base by half the 
sum of the greatest and least lengths of the ungula. 

I. What is the convex surface of the cylindric angula A KB A 
NM, if ^4^Vis8 feet, BM 12 feet and the radius BE of the 
base 3 feet? 

By formula, S=$(a-\-a')27rr=7r(a+a') r=7e(S+12) X3=* 
188 49552 sq. ft. 

1. 8 ft the least length A JV of the ungula, and 

2. 12 ft.=the greatest length BM. 

3. 10ft.=J(8ft.+12 ft.)=half the sum of the least aua 

greatest lengths. 

4. 18.849552 ft.=67r=the circumferenc of the base. 

5. .-. 10X18.849552=188.49552 sq. ft.=the convex surface, 



258 FINKEL'S SOLUTION BOOK. 

III. .'. 188.49552 sq. ft.=the convex surface of the ungula. 

Prob. LXXXIII. To find the volume of a cylindric ungula, 
when the plane passes obliquely through the opposite sides 
of the cylinder. 

Formula. V=a-a / 7 tr' 1 



Multiply the area of the base, by half the least and 
greatest lengths of the ungula. 

I. What is the volume of a cylindric ungula whose least 
length is 7 feet, greatest length 11 feet, and the radius of the 
base 2 feet? 

By formula, K=J(+')7rr 2 = i(7+ll)7r2 2 =113.097312 cu. ft. 



II. 



1. 7 ft.=the least length of the ungula, and 

2. 11 ft.=the greatest length. 

3. 9 ft.=J(7 ft.+ll ft.)=half the length of the least and 

greatest lengths. 

4. 12.566368 sq. ft.=7r2 2 =the area of the base. 

5. .'.9X12.566368=113.097312 cu. ft.=the volume ot the 
H ungula. 



III. .-. The volume of the ungula is 113.097312 cu. ft. 

Prob. L.XXXIV. To find the convex surface of a cylindric 
ungula, when the plane passes through the base and one of 
its sides. 



rdx 




(r )vers -1 - | 



Rule. Multiply the sine of half the arc of the base by the 
diameter of the cylinder, and from the product subtract the prod- 
uct of the arc and cosine; this difference multiplied by the quo- 
tient of the height divided by the versed sine 'will be the convex 
surface. 

I. What is the convex surface of the cylindric ungula A CB 



MENSURATION. 



259 



D, whose altitude BD is 28 feet, height BM of arc of base 4 
feet and chord A C 16 feet? 



By formula, 5=2r 



(r-i 



28 



, =2XlOX 



j [V2X10X4 4 2 ( 10 4) vers-'Aj , = 




367200' 
= = 140[8 6.5638]=341.068 sq. ft. 



1. 28 ft.=the altitude BD. 

2. 4 ft.=the height BM of the arc of the base. 

3. 16 ft.=the chord A C of the arc of the base. 

4. 8 ft.=the sine CM of the arc. 

5. 10ft (8 2 +4 2 )-h-(2x4)=the radius OC=OB of the 

base, by Prob. XX, formula /?=(0 8 +c 2 )-i-2a. 

6. 6 ft.=10 ft 4 ft.=cosine OM of the arc. 

7. 160 sq. ft.=20x8=sine multiplied by the diameter of 

the base. 



II. 



8. 18.5438 



l=the arc 



CBA, by formula of Prob. XXV. 

9. /. 111.2628 sq. ft.==6Xl8.5438=the arc multiplied by 
the cosine OM. 

10. 160 sq. ft 111.2628 sq. ft.=48.7372sq.ft.=the differ- 

ence. 

11. .-. 341.1604 sq. ft.=(28-5-4)X 48.7372 sq. ft=the con- 

vex surface. 

III. .-. The convex surface is 341.1604 sq. ft. nearly. 

NOTE. The difference in the two answers is caused by the length of the 
arc CBA, in the solution, only being a near approximation. 



* Demonstration. In the figure, let BK=x, BM=b, 
and the angle BMD=S. Then MK=bx, and IK=FL=MK 

^ 



tan 6=(b x) tan 8. But tan^==. .'. FL=(bx). 



Now if we take an elemennt of the arc LBH, and from it draw 
a line parallel to FL, we will have an element of the sur- 



260 FINKEL'S SOLUTION BOOK. 

face LBHEGF. This will be a rectangle whose length is 
~(b x) and width an element of the arc LBH. An ele- 



ment of the arc is ds=*(dx*+dy z ). Let HK=y. Then^ 2 
2rx # 2 , by a property of the circle, from which we find 



*_ y T'fLX 

dx. .'. //.?== .'. The area of the element of 




a 
the surface is -(b x ) /_. ^^^ and the whole surface of AB C 

^ ' 2 



x a /* 

,^= f lr-i 
x 2 Wo 





2 ( r & ) r vers- 1 . 



Prob, LXXXV. To find the volume of a cylindric ungula, 
when the cutting 1 plane passes throug-h the base and one of 
its sides. 

/& a /& _ 

V= I (bx)dA=- l I (l>x)2\2rxx 2 (tx, 

Jo vJo 



. When x=0, 
=0. .-. C= i;rr2 



Rule. From f of the cube of half the chord of the base, sub- 
tract the product of the area of the base and the difference of the 
radius of trie base and the height of the arc of the base; this dif- 
ference multiplied by the quotient of the altitude of the ungula by 
the height (versed sine} of the arc of the base, 'will give the vol- 
ume. 

I. What is the volume of a cylindric ungula, whose altitude 
BD is 8 feet, chord A C of base 6 feet, and height BMvl arc of 
base 1 foot? 

By formula, V=- \ f(2r 2 ) f (r b} [~ ^rr 2 



MENSURATION. 261 

-r* sin - =8 f (2X5X1-1 ) f 

18 



(5 l)[i-5 a 4V2X5X1 1 5 2 sin" 1 ^Q j , =8 j 



1225 sin- 1 !"! I =528+800 sin' 1 ~200*= 
o J ) o 

13.20394 cu. ft. 

1. 8 ft=the altitude 

2. 1 ft.=the altitude BM of the arc ABC of the base. 

3. 6 ft.=the chord A C of the base. 

4. 18 cu. ft= f of 3 3 =| of the cube of the sine of half the 

arc of the base. 
I 3 

n.< " 

ula, (J), Prob. XXVIII. 

6. 16 cu. ft.=4 X 4 T V=the area of the base X OM, the cosine 

of the arc CHB. 

7. /. 8(18 cu. ft 16 cu. ft.)=13i cu ^ ft.=the volume of 

the cylindric ungula A CB D. 
III. .\ The volume of the cylindric ungula A CB D is 134 
cu. ft., nearly. 

Prob. LXXXVI. To find the convex surface of the frustum 
of a cylindric ung-ula. 



5. 4 T V sq. ft= f r o+fof 6Xl=area of the base, by form- 



Formula. S= 



Rule. ( 1 ) Conceive the section to he continued, till it meets 
the side of the cylinder produced; then say, as the difference of 
the heights of the arcs of the two ends of the ungula, is to the 
height of the arc of the less end, so is the height of the cylinder to 
the part of the side produced. 

(2) Find the surface of each of the ungulas, thus formed, 
fy Prob. LXXXIy., and their difference will be the convex sur- 
face of the frustum of the cylindric ungula. 

Prob. LXXXVII. To find the volume of a frustum of a 
cylindric uiig*ula. 

Formula. K=i(2rJ 3 2 ) f (r b) 1^ 2 



262 FINKEL'S SOLUTION BOOK. 



. Find the volume of the ungula whose base is the the 
upper base of the frustum and altitude that as found by (1) of the 
last rule. Also the volume of the ungula whose base is the lower 
base of the frustum and altitude the sum of the less zingula and 
altitude of the frustum. Their difference will be the volume of 
the frustum. 

3. PYRAMID AND CONE. 

Prob. LXXXVIII. To find the convex surface of a right 
cone. 

Formula. S= Cxi/*=27Tr x %Va 2 -{-r* ,. where C is the 
circumference, h the slant height, r the radius of the base, and a 
the altitude. 

Rule. Multiply the circumference of \ the base by the slant 
height and take half the product. Or, if the altitude and radius 
of the base are given, multiply the circumference of the base ly the 
square root of the sum of the squares of the radius and altitude, 
and take half the product. 

I. What is the convex surface of a right cone whose altitude 
is 8 inches and the radius of whose base is 6 inches?. 



By formula, S= 2^rXiVa 2 +r 2 
188.495559 sq. in. 

1. 6 in.=the radius AD of the base, 

and 

2. 8 in. the altitude CD. 

3. 10 in.=V r 8 2 +^ 2 =the slant 

height CA. 
11.14. 37.6991118 in.=2frr=12x 

3.14159265=the circumfer- 
ence of the base. 

4. .-. 188.495559 sq. in= 

i(10x37.6991118)=the con- 

vex surface of the cone. p,Q 

III. .'. The convex surface of the cone is 188.495559 sq. in. 

Prob. LXXXIX. To find the convex surface of a pyramid. 

Formula. S=$pXb, in which p is the perimeter of 
the base and h the slant height. 

Rule. Multiply the perimeter of the base by the slant height 
and take half the product. 




MENSURATION. 26a 

I. What is the convex surface of a pentagonal pyramid whose 
slant height is 8 inches and one side of the base 3 inches? 

By formula, fc^/ X/*=i( 3+3+3+3+3 ) X 8=60 sq. in. 

1. 8 in.=the slant height. 

2. 3 in. the length of one side of the base. 
II..J3. 5X3 in. =15 in.=the perimeter of the base. 

4. .'. 1(15X8)=60 sq. in.=the convex surface of the pyra- 

mid. 
III. .*. The convex surface of the pyramid is 60 sq. in. 

Remark. If the entire surface of a pyramid or cone is required, 
to the convex surface add the area of the base. 

Formula. T=S-\-A, where A is the area of the base 
and S the convex surface. 

Prob. XC. To find the volume of a pyramid or a cone. 

Formula. V=^aA=^aX 7 tr 2 , where a is the altitude 
and A=7tr 2 the area of the base. 

Rule. Multiply the area of the base by the altitude and take 
one-third of the product. 

I. What is the volume of a cone whose altitude CD is 18 
inches and the radius AD of the base 3 inches? 



By formula, V=^a X?rr 2 =4x 18 X 7r3 2 =54x 3.14159265= 
169.646 cu. in. 

1. 18 in.=the altitude CD, and 

2. 3 in.=the radius AD. 



II. 



3. 28.27433385 sq. in.=z:7rr 2 =3 2 7r=the area of the base. 



4. .'. 169.6460031 cu. in.=n^4=xl8x3 8 7r=the volume 

of the cone. 
III. .-. The volume of the cone is 169.6460031 cu. in. 

Prob. XCI. To find the convex surface of a frustum of a 
cone. 

Formula. S= 



+ (r r') 2 , in which C is the circumference of the 
lower base, C x the circumference of the upper base, and ^,= 
Va*+(r r'}* , the slant height. 

Rule. Multiply half the sum of the circumferences of the 
two bases by the slant height. 

1. What is the convex surface of the frustum of a cone whose 
altitude is 4 feet, radius of the lower base 4 feet, and the radius of 
the upper base 1 foot? 



264 FINKEL'S SOLUTION BOOK. 



By formula, 5=7i'(r+r / )Va 2 +(r~r / )2=^(4+l)\ / 4 2 +(4 
.=257r.=78.539816 sq. ft. 

1. 4 ft.= the altitude OE, 

2. 4 ft.=the radius AE of the lower 

base, and 

3. 1 ft.=the radius DO of the upper 

base. 

4. 3 ft.=^4^ PE(=DO)=r r>. 

5. 5 ft= 




=AD, the slant height. 

6. 87r=the circumference A GJBffot 

the lower base. 

7. 2w=the circumference DIC of the upper base. 

8. 57r=i(8 7 r-{-27r )=half the sum of the circumferences. 

9. .-. 5x57r=257T=78.539816sq. ft.=the convex surface of 

the frustum. 
III. .-. The convex surface of the frustum is 78.539816 sq. ft. 

Remark. If the entire surface of the frustum is required, to 
the convex surface add the area of the two bases. 

Formula. T=S+A+A / =7r(r+r / )*l a *+(rr')* + 



Prob. XGII. To find the convex surface of the frustum of 
a pyramid. 

Formula. S= 



Rule. Multiply half the sum of the perimeters of the two 
bases ~by the slant height. 

I. What is the convex surface of the frustum of a pentagonal 
pyramid, if each side of the lower base is 5 feet, each side of the 
upper base 1 foot, and the altitude of the frustum 10 feet? 

Before we can apply the formula, we must find the slant 
height. Produce FO, till OK=OE. Divide OK into extreme 
and mean ratio at H. Draw EH. Then KO : OH'. : OH : KH. 



whence OH*+KOx OH^KO*. Completing the square of this 
equation, OH*+KC>X OH+lKQt^KO* , from which OH(= 



MENSURATION. 



265 



. But EF= 




2V ; 5), and te 
where 5 is a side of the lower hase,= 



m ay be considered the radius R of a circum- 
VlO 2V5* 

scribed circle of the lower base. In like manner, the radius r of 
the circumscribed circle of the upper base may be found to b 

__Jfl , where s' is a side of the upper base,- ==. 
VlO 2V5 V5 



the apothem of the lower base,^ 




=|V650+lWl==the slant height. 



II. 



By formula, S=(25+5; 

155.5795 sq. ft 

1. 10 ft.=the altitude oO. 

5 fa = EA, one of the equal sides < 

the lower base. 
1 K.=ed 9 one of the equal sides of 

the upper base. 

4. f V650+W1=/^ the slant height. 
5X5 ft.=25 ft.=the perimeter of the 
lower base. 

6. 5X1 ft.=5 ft.=the perimeter of 

upper base. 

7. .-. -H 25+5 )|V650+10Vl = 155.5795 sq.ft.=the conve, 

surface. 

III. /. The convex surface of the frustum is 155.5791 
Prob. XCIII. To find the volume of a frustum of apyia- 
mid or a cone. 




266 FINKEL'b SOLUTION BOOK. 

Formula. (a) V=%a(A+VAA~'+A'), in which A 
is the area of the lower base, A' the area of the upper base and 
V ' AA' the area of the mean base. When we have a frustum of 
a cone, (b) V=$a(A+fAA'+A')=$a( 7tR*+V( xR* X nr* )+ 




Rule. (1) Find the area of the mean bast by multiplying the 
area of the upper and lower bases together and extracting the 
square root of the product. 

(2) Add the upper, lower, and mean bases together and multi- 
ply the sum by -J the altitude. 

I. What is the solidity of a frustum of a cone whose altitude 
is 8 feet, the radius of the lower base 2 feet, and the radius of 
the upper base 1 foot? 

By formula (3), V^Tta^+Rr+r*)^* 8(4+2+1)= 
X 56 TT =58.6433 cu. ft. 

1. 8 ft=the altitude. 

2. 2 ft.=the radius of the lower base. 

3. 1 ft.=the radius of the upper base. 

4. 4^=the area of the lower base. 

5. 7r==the area of the upper base. 

6. 27r=V47r X 7f=the area of the mean base. 

7. 4 7t + 7T+2 n==l 7t =the sum of the areas of the three bases. 
.'. itX8x7# =58.6433 cu. ft.=the solidity of the frustum, 

III. .-. Ths solidity of the frustum is 58.6433 cu. ft 

4. CONICAL UNGULAS. 

1. A Conical Unyula ( Lat. ungula, a claw, hoof, from 
unguis, a nail, claw, hoof) is a section or part of a cone cut ofl 
by a plane oblique to the base and contained between this plane 
and the base. 

Prob. XCIV. To find the surface of a conical imgula. 
Formula. S= 



II, 




i-(2/? t\r (^?+r t)x~\ ) , 
2* cos" 1 ' ^ ' I > dx, where a is the altitude 



of the ungula, R the radius of the base, ^the radius of the upper 
base of the frustum from which the ungula is cut, / the distance 
the catting plane cuts the base from the opposite extremity of 
the base, and x the radius of a section parallel to the base and at 
a distance h y from the base. 



MENSURATION. 267 

I* rob. XCV. To find tlie volume of a conical uiigula. 



C R 
Formula. F= / A dy= 



(Rr)x 



t)(R+r t) 



dx, 

where the letters represent the same value as in the preceding 
problem and dy=( -- \dx, since y= ^ -- -. 

^L \. "?* J _/\ ' /* 

Prob. XCVT. To find the convex surface of a conical un- 
JMI la, when the cutting 1 plane passes through the opposite 
extremities of the ends of the frustum. 



Formula. S= 



This formula is obtained by putting t 0, in 
the formula of Prob. XCIV., and integrating the 
result. For, in this problem, the cutting plane 
AHCK passes through the opposite point A, and 
therefore the distance from A to the cutting 
plane is 0. .-. t0. 




FIG. 43. 



Rule. Multiply half the sum of the radii of the bases by the 
square root of their -product and subtract the result from the 
square of the radius of the lower base. Multiply this difference 
by 7t times the slant height and divide the result thus obtained by 
the difference of the radii of the bases. 

Prob. XCVII. To find the volume of a conical ungrula, 
when the cutting- plane passes through the opposite extremi- 
ties of the ends of the frustum. 



Formula.- l^^^, (&-**) 
3(/c r) V J 



This formula is obtained by putting t=0, in the formula of Frob. 
XCV., and integrating the result. 



. Multiply the difference of the square roots of the 
cubes of the radii of the bases by the square root of the cube of the 
radius of the lower base and this product by \n times the altitude. 



268 FINKEL'S SOLUTION BOOK. 

Divide this last product by the difference of the radii of the two 
bases and the quotient 'will be the volume of the ungula. 

I. A cup in the form of a frustum of a cone is 7 in. in diame- 
ter at the top, 4 in. at the bottom, and 6 in. deep. If when full 
of water, it is tipped just so that the raised edge of the bottom 
is visible; what is the volume of the water poured out? 



By formula, F^7? == |^( 49-8^7)= 
102.016989 cu. in. 

Remark. Fig. 43 inverted represents the form of the cup and 
APBQ C the quantity of water poured out, C being the tipped 
edge of the bottom. 

I. A tank is 6 feet in diameter at the top, 8 feet at the bot- 
tom, and 12 feet deep. A plane passes from the top on one side 
to the bottom on the other side : into what segments does it 
divide the tank? 

By formil ,a, K = 
32 7 r(8-3V / 3)=281.87 cu ft. 



II. 



1. 4 ft.= AL, the radius of the lower base. 

2. 3 ft.=Z?/ ?1 , the radius of t"he upper base, and 

3. 12 ft.=^Z, the altitude. Then 

_V3" 8 )= 327r(8-3\/3)=281.87 



3(4-3) 
sq. ft. the volume. 

III. .'. The volume is 281.87 cu. ft 

Prob. XCVIII. To find the convex surface ofa conical un- 
gula, when the cutting- plane FCE makes an angle CIB less 
than the angle DAB, i. e. when AIi=t) is less thanDC(=2r). 



Formula. S= 




This formula is obtained by integrating the formula of 
Prob. XCIV, recollecting that the co- efficient of x 2 is negative. 

Prob. XCIX. To find the volume of a conical ungrula, when 
the cutting' plane FCE makes an angle CIB less than the an- 
gle DAB. i. e., when AI (=t) is less than CD (^3r). 



MENSURATION. 



269 



Formula.- V= 




FIG. 44. 



This formula is obtained by integrating the 
formula of Prob. XCV, recollecting that the co- 
efficient of x 2 is negative. 

Prob. C. To find the convex surface of a conical un- 
gula, when the cutting- plane FCE is parallel to the side AD, 
i. e., when AI(=t) is equal to DC(=2r). 



Formula. S== 



cos 



-r)r-%(R-r^(R r}r\ 



This formula is obtained by putting /=2r, 
in the formula of Prob. XCIV., and inte- 
grating the resulting equation. 

Fob. CI. To find the volume of a coni- 
cal uiig-ula, when the cutting- plane FCE 
is parallel to the side DA, i. e., when 
AI(=t) is equal to CD (=2r). 



Formula.- V 




FIG 45. 



This formula is obtained by putting /=2r, in the formula of 
Prob. XCV., and integrating the resulting equation. 

Prob. CII. To find the con vex surface of a concial ung-ula, 
when the cutting plane FCE makes an angle C1B greater 
than the angle DAB, i. e., when AI (=t) is greater than DC 



Formula.- 



270 FINKEL'S SOLUTION BOOK. 

This formula is obtained by integrating the formula of Prob. 
XCIV., remembering that the coefficient of x 2 , which occurs in 
process of integrating, is positive. 

Prob. CHI. To find the volume of a conical nngula, when 
the cutting plane FCE makes an angle CIB greater than the 
angle DAB, i. e., when AI(=t) is greater than DC(=2r). 



Formula 




This formula is obtained by integrating the formula of Prob, 
XCV., regarding the coefficient of* 2 positive. 



XII. THE 
Prob. CIV. To find the convex surface of a sphere. 



Formula. S=2 X 2^yV^>' 3 4-^ 2 =47r7? 2 = ?rZ> 2 , where 
Z? is the diameter. 

Rule. Multiply the square of the diameter by 3.14-1592. 
I. What is the surface of a sphere whose radius is 5 inches? 
By formula, S=7tR 3 =4^X25=314.1592 sq. in. 

II. 5 in.=the radius. 
2. 25 sq. in.= the square of the radius. 
3. /. 4#X25 sq. in.=314.1592 sq. in.=the surface of the 
^ sphere. 

III. .'. 314.1592 sq. in.^the surface of the sphere. 

NOTE. Since 7r/? 2 is the area of a circle whose radius Is /?, the area 
(4n-/? 2 ) of a sphere is equal to four great circles of the sphere. The sur- 
face of a sphere is also equal to the convex surface of its circumscribing 
cylinder. 

Proh. C V. To find the volume of a sphere, or a globe. 
Formula. V=&ity* '<**=* R*= f x(\D) 3 =| 



Rule. Multiply the cube of the radius by far (=4.188782); or 
multiply the cube of the diameter by \TT (=.5235987). 

I. What is the volume of a sphere whose diameter is 4 feet? 



MENSURATION. 271 

By formula, F=f7T./? 3 =|7r2 3 =33.510256cu. ft. 

1. 2 ft.=the radius. 

2. 8 cu. ft=2 3 =the cube of the radius. 

t<! 3. /. 4.188782X8 cu. ft.=33.510256 cu. ft.==the volume of 

the sphere. 

III. .-. 33.510256 cu. ft.=the volume of the sphere. 
Prob. CVI. To find the area of a zone. 

A. ZiOn,e Is the curved surface of a sphere included between two 
parallel planes or cut off by one plane. 

Formula. S=^nRa, in which a is the altitude of the 
segment of which the zone is the curved surface. 

Rule. Multiply the circumference of a great circle of the 
sphere by the altitude of the segment. 

I. What is the area of a zone whose altitude is 2 feet, on a 
sphere whose radius is 6 feet? 

By formula, 5=2*^=2 ?rQx 2=24 n =75.39822 sq. ft. 

1. 6 ft.=the radius of the sphere. 

2. 2 ft.=the altitude. 

II.<(3. 12 ?r=37. 69911 ft.=the circumference of a great circle of 

the sphere. 

4. .'.2X37.69911=75.39822 sq. ft.=the area of the zone. 
III. .-. The area of the zone is 75.39822 sq. ft. 

NOTE. This rule is applicable whether the zone is the curved surface of 
the frustum of a sphere or the curved surface of a segment of a sphere. 

Prob. CVII. To find the volume of the segment of a sphere. 

Formula. V=.\na( s &r\-\-a CL ) where r 1 is the radius of 
the base of the segment. 

Rule. To three times the square of the radius of the base, add 
the square of the altitude and multiply the sum by ^n=.52S5987 
times the altitude. 

I. What is the volume of a segment whose altitude is 2 inches 
and the radius of the base 8 inches? 

By formula, F=l7ra(3r;+*2)=-^X2(3X64+4)=205.2406 
cu. in. 

1. 8 in.=the radius of the base. 

2. 2 in.=the altitude of the segment. 

3. 192 sq. in = g 3x8 2 =three times the square of the radius. 



II. 



4. 4 sq. in.=the square of the altitude. 

5. 196 sq. in.=192 sq. in.-f4 sq. in. =three times the square 

of the radius plus the square of the altitude. 

6. |7TX2 X 196=205.2406 cu. in.=the volume of the segment- 



272 FINKEL'S SOLUTION BOOK. 

III. /. 205.2406 cu. in =the volume of the segment. 



N OTE . From the formula F=^a(3r-(- rt2 )> we nave V \-xa 
But \xar\ is the volume of a cylinder whose radius i$r lt and altitude {a, and 
i7r3 is the volume of a sphere whose diameter is a :. The volume of a 
segment of a sphere is equal to a cylinder whose base is the base of the seg- 
ment and altitude half the altitude of the segment, plus a sphere whose 
diameter is the altitude of the segment. 

Prob. CVIII. To find the volume of a frustum of a 
sphere ,or the portion included between two parallel planes. 

Formula. V^7ta\Z(rl+rl)+a*-\=\a(nrl+nr\) + 
\7t a* *, in which r l is the radius of the lower base, r 2 the ra- 
dius of the upper base. 

Rule. To three times the sum of the squared radii of the two 
ends, add the square of the altitude; multiply this sum by .5235987 
times the altitude. 

I. What is the volume of the frustum of a sphere, the radius of 
whose upper base is 2 feet and lower base 3 feet and altitude -J- 
foot? 

By formula, V=\ *a[3(rH-rJ)+a*]=i,r xi[3(9+4)+i]= 
8.03839 cu. ft. 

1. 3 ft. the radius of the lower base. 

2. 2 ft. the radius of the upper base. 

3. 39 sq. ft.=--3(3 2 -f2 2 )=three times the sum of the squares 



II. 



of the radii of the two bases. 



4. \ sq. ft.=the square of the altitude. 

5. .-. |TT XiX39^=8.03839 cu. ft=the volume of the 

frustum. 

III. .-. 8.03839 cu. ft.=the volume of the frustum. 
Prob. CIX. To find the volume of spherical sector. 

A Spherical Sector is the volume generated by any sector 
of a semi-circle which is revolved about its diameter. 

Formula. l / =^7ta7? 2 , where a is the altitude of the 
zone of the sector. 

Rule. Multiply its zone by one-third the radius. 

*NOTK. rt(?r ;?-(- Trr j )=the volume of two cylinders whose bases are 
the upper and lower bases of the segment and whose altitude is hall the alti- 
tude oi the segment. ^Tra 3 is the volume of a sphere whose diameter is the 
altitude of the segment. Hence the volume of a segment of a sphere of two 
bases is equivalent to the volume of two cylinders whose bases are the up- 
per and lower bases respectively of the segment and whose common altitude 
is the altitude of the segment, plus the volume of a sphere whose diameter 
is the altitude of the segment. 

For a demonstration of this and the preceding formula, see Wenttvorth's 
Plane and Solid Geometry, Bk. IX., Prob. XXXII. 



MENSURATION. 



273 



I. What is the volume of a spherical sector the altitude of 
whose zone is 2 meters and the radius of the sphere 6 meters? 

By formula, F=f jw/? a =fjr X2x6 2 = 
150.7964m 3 . 

"1. 2m.=the altitude BD of the zone gener- 
ated by the arc EF when the semicir- 
cle is revolved about AB. 

2. 6m.=the radius EC of the sphere. 

3. 2 7r6m.=37. 699104 m =the circumference 

of a great circle of the sphere. 

4. 2 7t 6 X 2=75.398208 m 2 .=the area of the 

zone generated by EF, by Prob. CVI. 

5. .-. ix6x75.398208=150.796416ms=the volume of the 

spherical sector. 

Ill .-. The volume of the spherical sector is 150.796416 m 8 . 

I. Find the diameter of a sphere of which a sector contains 
7853.98 cu. ft., when the altitude of its zone is 6 feet. 



II., 




FIG. 47. 



By formula, V= 
7853.98 cu. ft, or 4r 2 =2500 sq. ft., whence 2r=50 feet, the 
diameter of the sphere. 

1. 6 ft.=the altitude of the zone. 

2. .'. \ir. x6X?' 2 =the volume of the sector. But 

3. 7853.98 cu. ft=the volume. 

4. .-. |7rx6X^ 2 ==7853.98 cu. ft. 

5. r 2 =625 sq. ft, by dividing by 4;r. 

6. .'. 2r=50 ft., the diameter of the sphere. 

III. .-. The diameter of the sphere is 50 feet. 
Prob. CX. To find the area of a lune. 



II. 



A. Lime is that portion of a sphere comprised between two 
great semi-circles. 

* f A 

Formula. S=^7tR^{ 

the quotient of the angle of the lune divided by 360. 



where u is 



. Multiply the surface of the sphere by the quotient of 
the angle of the lune divided by 360 

I. Given the radius of a sphere 10 inches; find the area of a 
lune whose angle is 30. 



By formula, S=7t R* u=4X * XlO 2 X(30-f-360 )= 
7rl0 2 =104.7197 sq. in. 



274 FINKEL'S SOLUTION BOOK. 

1. 10 in.=the radius of the sphere. 

2. 30=the angle of the lune. 

3. T 1 T p=30-r-360=:the quotient of the angle of the lune 



II. 



divided by 360. 

4. 47rl0 2 =400^=1256.6368 sq. in.=the suiface of the 

sphere. 

5. .'. T VX 1256.6368 sq. in.=104.7198 sq. in.=the area of the 

lune. 
III. .-. The area of the lune is 104.7198 sq. in. 

Went-wortW s New Plane and Solid Geometry, p. 371, Ex. 585. 
Prob. CXI. To find the volume of a spherical ungula. 



A. Spherical U^ngula is a portion of a sphere bunded by a 
lune and two great semi-circles. 

Formula. V=\n R*u, where u is the same as in the 
last problem. 

Rule. Multiply the area of the lune by one-third the radius; 
or, multiply the volume of the sphere by the quotient of the angle 
of the lune divided by 360. 

I. What is the volume of a spherical ungula the angle of 
whose lune is 20, if the radius of the sphere is 3 feet? 



By formula, V=$ * R* U=TT x 3 3 X(20-r-360) = 6.283184 
cu. ft. 



II. 



r l. 3 ft.=the radius of the sphere. 

2. 47r3 2 X(20-j-360>=6.2S3184s q . ft.=the area of the 
lune, by Prob CX 



3. .'. X3X6.283184=6.283184 cu. ft.=the volume of the 

ungula. 
III. .-. 6.283184 cu. ft. is the volume of the Ungula. 

Prob. CXII. To find the area of a spherical triangle. 

Formula. 3=2* R* x (A^-B+C 180)-r-360, in 
which A, B, and C are the angles of the spherical triangle. 

Rule. Multiply the area of the hemisphere in which the tri- 
angle is situated by the quotient of the spherical excess (the ex- 
cess of the sum of the spherical angles over 180^) divided by 360. 

I. What is the area of a spherical triangle on a sphere 
whose diameter is 12, the angles of the triangle being 82, 98, 
and 100 ? 

By formula, 5=2 nR* x(A+-\-C lSO)-r-360 =27r6 2 X 
(82+98 +100 



MENSURATION. 275 

1. 6=the radius of the sphere. 

2. 27r6 2 =727r=the area of the hemisphere. 

T ,3. (82+98+100 180)=100=the spherical excess. 

4. 100 -7-360 = T 5 =the quotient of the spherical excess 

divided by 360. 

5. .'. T 5 3X72nr=*62.83184=the area of the spherical triangle. 
III. .'. The area of the spherical triangle is 62.83184. 
(Olney*s Geometry and Trigonometry, Un. Ed., p .238, Ex. 8.) 

Prob. CXIII. To find the volume of a spherical pyramid. 

A Spherical Pyramid is the portion of a sphere bounded 
by a spherical polygon and the planes of its sides. 

Formula. F=f7r/? 8 X( ^-5-360), where B is the 
spherical excess. 

Rule. -Multiply the area of the base by one-third of the radius 
of the sphere 

I. The angles of a triangle, on a sphere whose radius is 9 feet, 
are 100, 115, and 120 ; find the area of the triangle and 
the volume of the corresponding spherical pyramid. 

By formula, F=| n R* x ( .-7-360 }=\nR* K(A+B+C 
18G )-7-360 = f nr9 3 X (100 +115 +120 180 ) -j- 360 = 
'657.a771$6 cu. ft. 

1. 9 ft.=the radius of the sphere. 

2. 27r9 2 =the area of the hemisphere in which the pyramid 

is situated. 

3 (100 +115 +120 180) 155=the sperical ex- 
cess. 



II. 



4. fi=l 55 -r-36G=the quotient of the spherical excess 



divided by 360. 
5. .-. fix27r9 2 HBX ^9 2 =the area of the base of the pyra- 

mid. 
- iX9XliX27r9 2 =657.377126cu. ft.=the volume of 

the pyramid. 

III. .'. The volume of the spherical pyramid is 657.377126 
cu. ft. 

( Van Amringe's Da-vies* Geometry and Trigonometry, p. 278, 
Ex. 15. 

I. Fmd the area of a spherical hexagon whose angles are 96 , 
110, 128, 136, 140, and 150, if the circumference of a 
great circle of the sphere is 10 inches. 

Formula. S=<l7tR* [ T ~~ ( U ~ \ where T is 



the sum of the angles of the polygon and n the number of 
sides. 



276 FINKEL'S SOLUTION BOOK. 



By formula, 

(96 +110 +128 +136 +140 +150 (6 2)xl80)-5- 
360 = X(760 720 )^-360=i =1.7684 sq. in. 

7t 7t 

1. 5-7-'T=the radius of the sphere, since 2?rR 10 in.. 

2. 760=96 +110 +128 +136 +140 +150=the 

sum of the angles of the polygon. 

3. 760 (6 2)xl80=40=the spherical excess. 

4. ^=40 -r-360=the quotient of the spherical excess di- 
rr < videdby360. 

><5 X2 

5. 2^11 the area of the hemisphere on which the 

polygon is situated. 

(5 \ 2 
- 1 =4x50-f-JT=1.7684 sq. in. 
7t S 

III. /. The area of the polygon is 1.7684 sq. in. 
Wentworttts Geometry, Revised Ed., p. 374, Ex. 596. 



XIII. SPHENOID. 

1. A Spheroid is a solid formed by revolving an ellipse 
about one of its diameters as an axis of revolution. 

1. THE PROLATE SPHEROID. 

1. The Prolate Spheroid is the spheroid formed by re- 
volving an ellipse about its transverse diameter as an axis of 
revolution. 

Prob. CXIV. To find the surface of a prolate spheroid. 

Formulae. (a) S= 



, where 
& & 

^2Z_32 



a 
surface. 



=the eccentricity of the ellipse which generates the 



-- 



MENSURATION. 



277 



Rule. Multiply the circumference of a circle whose radius 
$s the semi -conjugate diameter by the semi -conjugate diameter in- 
creased by the product of the arc whose sine is the eccentricity into 
the quotient of the semi-transverse diameter divided by the eccen- 
tricity. 

I. Find the surface of a prolate spheroid whose transverse 
diameter is 10 feet and conjugate diameter 8 feet. 

5 . 



By formula (a), S=27rt>(l>-\- 

e 



rA 




|7r[48+100x. 6435053] 235.3064 sq. ft. 

fl. 25. 1327412=2 7r4=the circumference of a circle whose 
radius is the semi-conjugate diameter of the ellipse. 



II. < 



2. = 



=the eccentricity. 






3. 2 ^ 5 ft.=5 ft-i-f =the quotient of the semi-transverse diame- 

ter divided by the eccentricity. 

4. .6435053=the arc (to the radius 1) whose sine is f , or the 

eccentricity. 

5. 5.3625442 ft.=Yft. X .6435053= % 5 ft. X the arc whose 

sine is -| 

6. 9.3625442 ft.=4 ft+5.3625442 ft=semi - conjugate di- 

ameter increased by said product. 

7. .-. 235.3064 sq. ft.=9.3625442x25.1327412=the surface 

of the prolate spheroid. 

III. /. The surface of the prolate spheroid is 235.3064 sq. ft. 
Prob. CXV. To find the volume of a prolate spheroid. 

1) 2 /*a 

Formula. F= Cny*dx=7t I (a 2 x 2 )dx= 
J a 1 J _ a 

r I a 2 x # 3 =^7rb' 2 a, in which b is the semi-conjugate 

I I d 

diameter, and a the semi-transverse diameter. 

Rule. Multiply the square of the semi -conjugate diameter by 
the semi-transverse diameter and this product by ^n. 

I. What is the volume of a prolate spheroid, whose semi-trans- 
verse diameter is 50 inches, and semi-conjugate diameter 30 inches. 

By formula, r=|^ 2 =|^30 2 X 50=188495.559 cu. in. 



278 FINKEL'S SOLUTION BOOK. 

1. 30 in.=the semi -conjugate diameter, 

2. 50 in.=the semi-transverse diameter. 

3. 900 sq. in=the square of the semi-conjugate diameter. 

[. 45000 cu. in.=50x900=the squareof the semi-conjugate 

diameter by the semi- transverse diameter. 
5. . -4^45000=1X3.14159265X45000 cu. in.= 

188495.559 cu. in.=the volume of the prolate spheroid. 
III. .*. The volume of the prolate spheroid is 188495.559 cu. in. 

2. THE OBLATE SPHEROID. 

1. An Oblate Spheroid is the spheroid formed by revolving 
an ellipse about its conjugate diameter as an axis of revolution. 

Prob. CXVI. To find the surfae of an oblate spheroid. 

r a 

Formulae. (a) S= C27rxds=2 I 2; 

'-a 



q 




Prob. CX VII. To find the volume of an oblate spheroid. 



II l lie. Multiply the square of the semi-transverse diameter- 
by the semi. conjugate diameter and this product by ^ n. 

I. What is the volume of an oblate spheroid, whose trans- 
verse diameter is 100 and conjugate diameter 60? 



By formula, V=$7t a* 6=$ie5Q 2 X30=314159.265. 

1. 30=^ of 60= the semi-conjugate diameter. 

2. 50 -J of 100=the semi-transverse diameter. 

3. 2500=50 2 =the square of the semi-transverse diameter. 



II. 



4. 75000=30 X2500=the square of the semi -transverse di- 



ameter multiplied by the semi-conjugate diameter. 

5. /. \n X 75000=314159.265= the volume of the oblate 

spheroid. 
III. .-. The volume of the oblate spheroid is 314159.265. 

NOTE Since the volume of a prolate spheroid is ^b^a. We may write 
7r 2 a (7r2X2rt). But 7r^2x2 is the volume of a cylinder the radius of 
whose base is b and altitude 2a. .'. The volume of a prolate spheroid is | of 
the circumscribed cylinder. In like manner, it may be shown that the vol- 
ume of an oblate spheroid is of its circumscribed cylinder. 

The following is a general rule for finding the volume of a 
spheroid; Multiply the square of the revolving axis by the Jixed 
axis and this product by \it . 



MENSURATION. 



Prob. CXVIII. To find the volume of the middle frustum 
of a prolate spheroid, its length, the middle diameter, and 
that of either of the ends being given. 

CASE I. 

When the ends are circular, or parallel to the revolving axis. 
F= T Vn'(2Z> 2 +^ 2 )/, where D is the middle 



diameter CD, d the diameter ///of an end, and / the length of 
the frustum. 

Rule. To twice the square of the middle diameter add the 
square of the diameter of either end and this sum multiplied by 
the length of the frustum, and the product again by -^n, 'will 
give the solidity. 

I. What is the volume of the middle frustum HIGF of a 
prolate spheroid, if the middle diameter CD is 50 inches, and 
that of either of the ends ///or FG is 40 inches, and its length 
OK 18 inches? 

By formula, F= T 1 ^7r(2/? 2 +c/ 2 )/= T 1 Tr 7r(2x50 2 +40 2 )18= 
31101.767265 cu. in. 

1. 50 in.=the middle diame- 

ter CD. 

2. 40 in.=the diameter of ei- 

ther end as HI. 

3. 18 in.=the length OK of 

the frustum. 

4. 5000sq. in.=2x50 2 =twice 

II. < the square of the middle 

diameter. 

5. 1600sq. in.=40 2 =the 

sqaureof the diameter of either end. 

6. 5000 sq. in.+1600 sq. in.=6GOO t.q. in. 

7. 18X6600=118800 cu. in. 

8. .'. T VwXH8800cu. in.=31101.767265 cu. in.=the vol- 

ume. 

III. .-. The volume of the frustum is 31101.767265 cu. in. 

CASE II. 

When the ends are elliptical, or perpendicular to the revolving 
axis. 




. V=^n(ZDd+D'd')l, where D and d are 
the transverse and conjugate diameters of the middle section and 
D / and d / the transverse and conjugate diameter of the ends and 
/ the distance between the ends. 

Rule. ( 1 ) Multiply twice the trans-verse diameter of the 
middle section by its conjugate diameter, and to this product add 



280 FINKEL'S SOLUTION BOOK. 

the product of the transverse and conjugate diameter of either of 
the ends. 

(2) Multiply the sum, thus found, by the distance of the ends, 
the height of the frustum, 
the result will be the volume. 



or the height of the frustum, and the product again by ^7t 
the 




I. What is the volume of the middle frustum of an oblate 
spheroid, the diameter of the middle section being 100 inches 
and 60 inches; those of the end 60 inches and 36 inches ; and the 
length 80 inches? 

By formula, V=^7t(%Dd-\-D'd') /=^w(2x 100x60+60 X 
36) 80=296566.44616 cu. in. 

1. 100 in.=the transverse diameter PC of the middle section. 

2. 60 in. =the conjugate diameter ms of the middle section. 

3. 12000sq. in.==2XlOOX60==twice the product of the 

diameters of the middle section. 

4. 60 in.=the transverse diameter 

AB of the end. 

5. 36 in.=the conjugate diameter 
r-r J %(nc) of the end. 

6. 2160 sq. in.=the product of the 

diameters of the end. 

7. 14160 sq. in.=12000 sq. in.+2160 

sq. in. 

8. 80X14160=1132800 cu. in. = the FIG. 49. 

product of said sum by the height of the frustum. 

9. .'. T V*X 1132800 cu. in.=296566.44616 cu. in.=the vol- 

ume of the frustum. 

III. .-. The volume of the frustum is 296566.44616 cu. in. 

Prob. CXIX. To find the volume of a segment of a prolate 
spheroid 

CASE I. 
When the base is parallel to the revolving axis. 

Formula. V=\nh*(--\' (ZDWi], where h is the 

height of the segment, d the revolving axis, and D the fixed 
axis. 

Rule. (1) Divide the square of the revolving axis by the 
square of the fixed axis, and multiply the quotient by the differ- 
ence between three times the Jixed axis and twice the heigat of 
the segment. 

(2) Multiply the product, thus found, "by the square of the 
height of the segment, and this product by \TT, and the result will 
be the volume of the segment. 



MENSURATION. 281 

I. What is the volume of a segment of a prolate spheroid of 
which the fixed axis is 10 feet and the revolving axis 6 feet and 
the height of the segment 1 foot? 



By formula, V=\nh* (%D 2h)= 



5.277875652 cu. ft. 

1. 10 ft.=the transverse diameter 

1BF. 

2. 6 ft.=the conjugate diameter AE. 

6 2 ' A 

3. 9 ^= =the square of the conju- 



II. 




gate diameter divided by the 
square of the transverse diameter. 

4. 28ft.=3XlO ft 2X1 ft.= the difference between three 

times the transverse diameter and twice the height of 
the segment. 

5. / 7 X28 ft.=10& ft.=the product of said quotient by 

said difference. 

7. .-. iVxlOA- cu. ft.=5.277865652 cu. ft.=the volume. 



III. /. The volume of the segment is 5.277875652 cu. ft. 

CASE II. 
When the base is perpendicular to the revolving axis. 

Formula. V=7r/i 2 (--.J(3d <2k), where d is the 
revolving axis, D the fixed axis, and h the height of the segment. 

Rule. (1) Divide the fixed axis by the revolving axis, and 
multiply the quotient by the difference between three times the 
revolving axis and twice the height of the segment, 

(2). Multiply the product, thus found, by the square of the 
height of the segment, and this product again by \rc. 

I. Required the volume of the segment of a prolate spheroid, 
its height being 6 inches, and the axes 50 and 30 inches respect- 
ively. 

By formula, V<^tk*() (3<*-2>&)=^x6 2 (|J)x 




282 FINKEL'S SOLUTION BOOK 

,(3X30 2X6)=2450.442267 cu. in. 

1. 50 in.=the transverse diameter, or 

axis. 

2. 30 in.=the conjugate diameter 2MO. 

3. !=50-=-30=the quotient of the trans 

verse diameter divided by the 

conjugate diameter. FIG. 57. 

4. 78 in.=3x30 in. 2x6in.=the difference between three 

times the conjugate or revolving axis, and twice the 
height of the segment. 

5. 130in.=|x78 in.=the product of said quotient by said 

difference. 

6. 4680 cu. in=130x6 2 =the square of the height of the 

segment by said product. 

7. .'. tfX4680 cu. in.=2450.442269 cu. in.=the volume ot 

segment. 

.III. .-. The volume of the segment is 2450.442269 cu. in. 

XIV. CONOIDS. 

1. A Conoid is a solid formed by the revolution of a conic 
section about its axis. 

I. THE PARABOLIC CONOID. 

1. A Parabolic Conoid is the solid formed by Devolving a 
parabola about its axis of abscissa. 

Prob. CXX. To find the surface of a parabolic conoid, or 
paraboloid. 

Formulae. (a) S 




where 2/ is the latus rectum of the parabola and y the radius of 
the base of the conoid, or the ordinate of the parabola. 

(3) 5 fTTV^j (p-\-*)* p*\j where 2^ is the same as above 

and x the altitude of the conoid, or the axis of abscissa of the 
parabola. 

Rule. To the square of half the latus rectum, or principal 
parameter, add the square of the radius of the base of the conoid 
and extract the square root of the cube of the sum; from this re- 
sult, subtract the cube of half the latus rectum and multiply the 



MENSURATION. 



difference ~by2n, and divide the product by one and one- half times 
the latus rectum. 

I. Determine the convex surface of a paraboloid whose axis is 
20, and the diameter of whose base is 60. 

From the equation of the parabola, y z = 
we have30 2 =2/x20; whence 2/=45. 

. By 




|n-X25x(125 27)=49X25X3.14159265= 
3848.45118. 



FIG. 52. 



II. 



1. 30= the radius AO of the base of the conoid. 

2. 20=the altitude OD. Then by a property of the para- 

bola, 

3. 30 2 =2/X 20, whence 

4. p=22$, the principal parameter of the parabola. 

5. 



=the square root of 

the cube of the sum of the squares of half the latus 
rectum and the radius of the base. 

6. 1 1 =the cube of half the latus rectum. . 



ence between said square root and the cube of half 
the latus rectum. 



o n vx m / -| o c o>7 \ TrN/QSVl ~" I ^TT timPS 

said difference. 
9. .'. 2^98x(-^- > ) -HlX45)=3848.45118=the surface of 

the conoid. 

III. .-. The surface of the conoid is 3848.45118. 
Prob. CXXI. To find the volume of a parabolic conoid. 



altitude. 



Formula. V=Jrty'*dx=j7t f lp x 
)x%7ry 2 x, whereby is the radius of 



the base and x the 



Rule. Multiply the area of the base by the altitude and take 
half the product. 



284 FINKEL'S SOLUTION BOOK. 

T. What is the volume of parabolic conoid, the radius of 
whose base is 10 feet and the altitude 14 feet? 

By formula, V=\n y* x = %n\W X 14=700x^=2202. 114855 
cu. ft. 

1, 10 ft.=the radius of the base. 

2. 14 ft.^the altitude. 

II.<3. 7rl0 2 =314.159265sq. ft. the area of the base. 

( /. |X14X314.159265=-2202.114855 cu. ft.=the volume 
of the conoid. 

III. .-. The volume of the conoid=2202.114855 cu. ft. 
NOTE. Since the volume of the conoid is %Ky 2 x, it is half of its circum- 
scribed cylinder. 

Prob. CXXII. To find the convex surface of a frustum of 
a parabolic conoid of which the radius of the lower base is R 
and the upper base r. 

In ( 3 a 

Formula. < ti 22 



rR n 

. S=J <2 l 7tyds=jj ) 



I. What is the volume of the frustum of a parabolic conoid of 
which the radius of the lower base is 12 feet, the radius of the 
upper base 8 feet, and the altitude of the frustum 5 feet? 

Since 12 2 =2/* / and 8 2 =2px, 12 2 S z =2p(x / x). Bnt*' x 
=5 feet. /. 12 2 8 2 =2/x5, whence 2/=16, the latus rectum. 

.-. By formula, sJ- 



Prob. CXXIII. To find the volume of the frustum of a 
parabolic conoid, when the bases are perpendicular to the 
axis of abscissa. 

Formula. V=^7tR z x / \7rr^x=^7i(x / 



Rule. Multiply the sum of the squares of the radii of the 
two bases by n and this product by half the altitude. 

I. What is the volume of the frustum of a parabolic conoid, 
the diameter of the greater end being 60 feet, and that of the 
lesser end 48 feet, and the distance of the ends 18 feet? 



By formula, V=%7ra(R*+r*) 
-f-576)==9Xl476X7T=13284T==41732.9177626 cu. ft, 



MENSURATION. 



285 



II 



III 



1. 30 ft. the radius of the larger base. 

2. 24 ft.=the radius of the lesser base. 

3. 18 ft.=the altitude of the frustum. 

4. 900 sq. ft.=the square of the radius of the lower base. 

5. 576 sq. ft. the square of the radius of the upper base. 

6. 1476 sq. ft =900 sq.ft.+576 sq. ft.^their sum. 

7. .-.-1X18X^X1476=13284X^=41732.9177626 cu. ft.= 

the volume ot the frustum of the conoid. 
/. The volume of the frustum is 41732.9177626 cu. ft. 
II. THE HYPERBOLIC CONOID. 

1. An Hyperbolic Conoid is the solid formed by revolv- 
ing an hyperbola about its axis of abscissa. 

Prob. CXXIV. To find the surface of an hyperbolic con- 
oid, or hyperboloid. 



Formula. S= 



^ ^ a / 

2* O-J- -4 dx=ln / - 
N x 2 a 2 J a 



=TT~ } x\e"x-a 2 - 
a ' 



a"~ 
e 



B-- 

a ( 



* a 2 ab\- 



-log 



ab 



1 ( 

e 

Prob. CXXV. To find the volume of an hyperbolic conoid. 

Formula. V=\n(R*-\-d' i )h, where 7? is the radius of 
the base, rtfthe middle diameter, and h the altitude. 

Rule. To the square of the radius of the base add the square 
of the middle diameter between the base and the vertex; and this 
sum multiplied by the altitude, and the product again by \7t, 
give the solidity. 

I. In the hyperboloid A CB, the altitude 
CO is 10, the radius A O of the base 12, 
and the middle diameter DE 15.8745; 
what is the volume? 

'l.'lO=the altitude CO. 

2. 12=the radius A O of the base. 

3. 15.8745=f he middle diameter DE. 

4. 144=12 2 =the square of the radius 
II J of the base. 

15. 251.99975=15. 8745 2 =the square of the middle diameter. 




FIG. 53- 



286 FINKEL'S SOLUTION BOOK. 

6. 395.99975=251.99975+144==the sum of the squares of 

the radius of the base and the middle diameter, 

7. .:. 7TXlQX395.99975=2073.454691=the volume. 
ill. ,-. The volatile of the conoid is 2073.454G91. 

Prob. CXXVI To find the volume of the frustum of an 
hyperbolic conoid. 



Formula. ^=$?T0(7? 2 -f-^ 8 + r *) where 7? is the ra- 
dius of the larger base, and r tne radius of the lesser base, and d 
the middle diameter of the frustum. 

Rule. Add together the squares of the greater and lesser 
semi -diameters, and the square of the whole diameter in the mid- 
dle; then this sum being multiplied by the altitude, and the prod- 
uct again by -J-7T, will give the solidity. 



QUADI^ATUr^E AND CUBAXUr<E OF 
SURFACES AND SOLIDS OF REVOLU- 
TION- 

1. CYCLOID. 

Prob. CXXVII. To find the surface generated by the 
revolution of a cycloid about its base. 



,S=2j27ryds=< 



\2ry 
Rule. Multiply the area of the generating circle by 6 ^. 

Prob. CXXVIII. To find the : volume of the solid formed 
by revolving' the cycloid about its base. 



Formula. 1= 




Rule. Multiply the cube of the radius of the generating cir- 
cle by 5?r 2 . 

Prob. CXXIX. To find the surface generated by revolv- 
ing" the cycloid about its axis. 



Formula. S=27ry^s=4:7ry=S7tr 2 ( n f ). 



Rule. Multiply eight times the area of the generating circle 
by n minus \. 

Prob. CXXX. To find the volume of the solid formed by 
revolving the cycloid about its axis. 



MENSURATION. 287 

Formulu. V= 



Rule. Multiply \ of the volume of a sphere whose radius is 
that of the generating circle by f /r 2 f . 

Prob. CXXXI. To find the surface formed by revolving 
the cycloid about a tangent at the vertex. 

Let Pbe a point on the curve, A=P=y J EP=A=x, 
A C= CF=r,and the angle A CF 
=6. Then \ve shall have AE= 
y = AC C=r r cos 6 ; and 
A B = x=FP+F= A F-\-EF 
_ r 0_|_ rs j n B. 

.-. Formula. S= 




(r F IG. 54. 



=I7tr* J 



llllle. Multiply the area of the generating- circle 

Prob. CXXXI 1 To find the volume formed by revolving 
a cycloid about a tang-ent at the vertex. 

Formula. V'^&Jny* ctx=<27rC 7r (rrcosff) 2 r(l+ 

(l cos0~ 

=7r 2 r 3 =the volume generated between the 
curve and the tangent. 



Rule. Multiply the cube of the radius of the generating cir- 
cle by 7zr 2 . 

2. CISSOID. . 

Prob. CXXXIII. To find the volume g-enerated by revolv- 
ing the cissoid about the axis of abscissa. 

Formula. Y 



288 



FLNKEL'S SOLUTION BOOK. 



Prob. CXXXIV. To find the volume formed by revolving 
the clssoid about its asymptote. 

Formula. V=2jit(AR)*dy(Fig. *j)=2* (2 *)*X 



( 2iCt> x j Y 

Prob. CXXXV. To find the volume formed by revolving 
the Witch of Agnesi about its asymptote. 



Formula. F / 7ty' 2 dx=\ ny 2 x 



Prob. CXXXVI. To find the volume formed by revolving 
the Conchoid of Nicomedes about its asymptote, or axis ol 
abscissa. 



Formula. V Cny^dx^^n I \ a _ 

J L_ V2 - 



2 



2. SPINDLES. 

A Circular Spindle is the solid formed by revolving the 
segment of a circle about its chord. 

Prob. CXXXVII . To find the volume of a circular spindle. 

Let AEBD be the circular spindle formed 
by revolving the segment A CBE about the 
chord A CB. Let AB=^a, the length of 
the spindle, and ED=^lb^ the middle diame- 
ter of the spindle. Let CI=K=x, the ra- 
dius of any right section of the spindle, and 

r=CZ,=y. Then the required volume of 



the spindle is 



r 

=2rt 
J o 



. ..(1). Let R= 




-f-24..(2), 



be the radius of the circle and 9 the angle A GB. Then by 
a property of the circle. KI^^^R El] y^EI, or y2 
El) Y.EI. But JEI=EGIG=R(IC+CG)=R 




whence x=*fi 2 y 2 Rcosd . . (3). Substituting this value of A 
in (1), we have V= 



MENSURATION. 



289. 



Rule. Multiply the area of the generating segment by tL 
path of its center of gravity, Guildin's Rule. 

3. THE PARABOLIC SPINDLE. 

A. IPuTdbolic Spindle is a solid formed by revolving a 
parabola about a double ordinate perpendicular to the axis. 

Prob. CXXXVIII. To find the volume of a parabolic 
spindle. 

Formula. V=2 C 7t(hxYdy= ( l 7 r f (/fc 2 

Jo Jo 



But x= 



Rule. Multply the volume of its circumscribed cylinder by r 8 ^. 

I. What is the volume of a parabolic spindle whose length 
A C is 3 feet and height BD 1 foot? 

By formula, F=if 7th*b=^7tX I 2 X 3=4.9945484 cu. ft. 
'1- 1 ft. height BD of the 
spindle. 

2. 3 ft=length A C. 

3. ?rXl 2 X 3=9.42477795 cu. ft. 

the volume of its circum- 
scribed cylinder. 
AX9.42477795 cu. ft.= 
4.9945484 cu. ft., the vol- 
ume of the parabolic spin- 
dle. FIG. 56. 
III. .'. The volume of the spindle is 4.9945484 cu. ft. 

Prob. CXXXIX. To find the volume generated by revolv- 
ing the arc of a parabola about the tangent at its vertex. 

Let A PC be an arc of a parabola revolved about AB, and let 
P be any point of the curve. Let AE=PF=x, and AF=PF 
==y t Then the area of the circle described by the line PF\& nx z . 

. Formula. V==. 



II. 



4. 




4^2 4/ 2 



290 FINKEL'S SOLUTION BOOK 

where /i=the height, and b= CD, the or- 



dinate of the curve. 

Rule. Multiply the volume of its circum- 
scribed cylinder by \. 

Prob. CXLi. To find the volume generated 
by revolving the arc APC of the parabola 
about BC parallel to the axis AD. 

The area of the circle generated by the line 
GP is 7t (b -yY"> 

FIG 57. 

.-. Formula. V=: 




Iillle. Multiply the volume of its circumscribed cylinder by \. 

NOTE. In the last two problems, the volume considered, lies between 
the curve and the lines AB and BC respectively. The volume generated 
by the segment ACD is found by subtracting the volume found in the two 
problems from the volume of the circumscribed cylinders. 

Prob. CXLJ. To find the volume formed by revolving a 
semi-circle about a tangent parallel to its diameter. 

Let the semi-circle be revolved about the tangent AG. Let 
A C=R , PF=A G=E C=y,AF= GP=x. Then 
the area of the circle generated by the line GP is 

Ttx*. But * 2 =2,ff 2 ZR(R Z >' 2 )^ y 2 ; for, 

; whence 
and X 2 =2fi 2 






Formula. V=2 JV 

-p _y 2 )dy=$# /? 3 ( 10 STT), which is the 

entire volume external to the semi-circle- 

FIG. 59. 

Rule. Multiply one-fourth of the volume of a sphere whose- 
radius is that of the generating semi-circle by (10 STT). 

XVI. TEOULAI SOLIDS. 



1. A Hef/ular Solid is a solid contained under a certain 
number of similar and equal plane figures. 

2. The Tetrahedron, or Regular Pyramid, is a 

regular solid bounded by four triangular faces. 

3. The Hexahedron, or Cube, is a regular solid bounded 
by six square faces. 

4. The Octahedron is a regular solid bounded by eight 
triangular faces. 

5. The Dodecahedron is a regular solid bounded by 
twelve pentagonal faces. 



MENSURATION. 



291 



6. The Icosahedron is a regular solid bounded by twenty 
equilateral triangular faces. 

These are the only regular solids that can possibly be formed. 

If the following figures are made of pasteboard, and the dotted 
lines cut half through, so that the parts may be turned up and 
glued together, they will represent the five regular solids. 




FIG. 59. 

1. TETRAHEDRON. 
Prob. CXLII. To find the surface of a tetrahedron. 

Formula. SW 2 V37where / is the length of a linear 



side. 



Multiply the square of a linear side by ^3=1.7320 



508. 

I. What is the surface of a tetrahedron whose linear edge is 
2 inches. 

By formula, S=/ 2 V3=2 2 V3=4V3==6.9282 sq. in. 
-1. 2 in.=the length of a linear side. 

side. [surface. 

.=6.9282 sq. in., the 
III. .-. The surface of the tetrahedron is 6.9282 sq. in. 

Prob. CXLIII. To find the volume of a tetrahedron. 



* i 

{1. 2 in.=the length of a linear side. 
2. 4 sq. in.=2 2 =the square of a linear side 
3. .-. V?X 4 sq. in.=1.73205x4 sq. m.==6.l 



side. 



Formula. F= T \V2 / 3 , where /is the length of a linear 



292 FINKEL'S SOLUTION BOOK. 

Rule. Multiply the cube of a linear side by T ^Vi, or .11785. 

I. Required the solidity of a tetrahedron whose linear side 
is 6 feet? 

By formula, V=^2 / 3 == T VV2x6 3 =18V2=25.455843 cu. ft. 



. 6 ft.=the length of a linear side. 

II. < 2. 216 cu. ft.^the cube of the linear side. 

3. .-. T 1 2V2"x216 cu. ft.=V2xl8 cu. ft.=25.45843 cu. ft. 

III. .-. The volume of the tetrahedron is 25.45843 cu. ft. 

2. OCTAHEDRON. 

Prob. CXLJV. To find the surface of an octahedron. 
Formula. 5=2^3 / 2 . 



5. Multiply the square of a linear side by 2*J~3 , i. e., by 
two times the square root of three. 

I. What is the surface of an octahedron whose linear side is 
4 feet? 

By formula, S=2*/3 / 2 =2V3X/ 2 =32V3=55.4256 cu. ft. 

1. 4 ft.=the length of a linear side. 

2. 16 sq. ft.:=4 2 the' square of the linear side. 

'<>. /. 2V3X16 sq. ft.=V3x32 sq. ft.=1.7320oX32 sq. ft.= 

55.4256 sq. ft. 
III. .-. The surface of the octahedron is 55.4256 sq. ft. 

Prob. CLXV. To find the volume of an octahedron. 

Formula. F=^V2 / 3 

Rule. Multiply the cube of a linear side by JV#, i. e., by one- 
third of the square root of two. 

I. What is the volume of an octahedron whose linear side is 
8 inches? 

By formula, V=#fa / 3 ^V2x8 3 ==.4714045X512==241.359104 
cu. in. 

1. g in. che length, of a linear side. 
]. 512 cu. in. 8 3 =the cube of a linear side. 
5. .-.iV / '2x512cu. in =^X 1-4142135X512 cu. in.= 
241. 359104 cu. in. 

III. . . The volume of the octahedron is 241.359104 cu. iu. 

3. DODECAHEDRON. 
Prob. CXLVI. To find the surface of a dodecahedron. 



MENSURATION. 293 



Formula. 5= 



Rule. Multiply the square of a linear side by 
, or 20.64.57285. 



I. What is the surface of a dodecahedron whose linear side is 
3 feet? 



By formula, 6-= 

185.8115565 sq. ft. 

. 3 ft.=the length of a linear side. 
2. 9 sq. ft.=3 2 =square of a linear side. 



IL 3. .-. 15x9 sq. ft=20.6457285x9 sq. ft. 



=185.8115565 sq. ft. 
III. The surface of the dodecahedron is 185.8115565 sq. ft. 

Prob. CXLVII. To find the volume of a dodecahedron. 



Formula. F= 



Rule 



. Multiply the cube of a linear side by 5+\{ - jjg -- V 

or 7.663115. 

I. The linear side of a dodecahedron is 2 feet ; what is its 
volume ? 



By formula, K= 
=61.20492 cu. ft. 

II. 2 ft.=the length of a linear side. 
2. 8 cu. ft.=2 2 =cube of a linear side. 
3. .-. 5V[A( 4 7+2lV5)]x8cu. ft.=7.663115x8cu. ft. 
=61.20492 cu. ft., the volume. 

III. .-. The volume of the dodecahedron is 61.20492 cu. ft. 

4. ICOSAHEDRON. 

Prob. CXI, VIII. To find the surface of an icosahedron. 
Formula. 5=5Vs/ 2 =8.66025 X / 2 . 



Rule. Multiply the square of a linear side by 5V^, or 
3.66025. 



294 



FINKEL'S SOLUTION BOOK. 



I. What is the surface of an icosahedron whose linear side 
is 5 feet. 

By formula, 5=5\ / 3/ 2 =5V3x5 2 =125V3=216.50625 sq. ft. 

(1. 5 ft.=length of a linear side, 
jj J2. 25 sq. ft.=5 2 =the square of a linear side. 

] 3. .'. 5^3X25 sq. ft.=8.66025x25 sq, ft.=216.50625 sq. ft. 

=the surface. 
III. /. The surface of the icosahedron is 216.5062^ sq. ft. 

Prob. CXLIX. To find the solidity of an icosahedron. 
Formula. ^=jV[i( 7+3V )] / 3 =2. 18169 X / 3 . 

'Rule. Multiply the cube of a linear side by fV[|(7+#VS)], 
or 2.18169 

I. What is the volume of an icosahedron whose linear side i& 
3 feet? 

By formula, F=|V[i(7+3V5)]/ 3 =2.18169 X3 3 =58.90563 cu.ft. 

II. 3 ft. the length of a linear side. 
2. 27 cu. ft.=:3 3 the cube of a linear side. 
3- .'. |V[i(7+3V5)]x27cu. ft.=2. 18169 X 27 cu. ft. 
=58.90563 cu. ft.=the volume. 

III. .-. The volume of the icosahedron is 58.905G3 cu. ft. 
NOTE. The surface and volume of any of the five regular sol- 
ids may be found as follows : 

Rule ( 1 ). Multiply the tabular area by the square of a linear 
side, and the product will be the surface 

Rule (2). Multiply the tabular volume by the cube of a 
linear side, and the product will be the volume. 

Surfaces and volumes of the regular solids, the edge being 1.. 



NO. OF 
SIDES. 


NAMES. 


SURFACES. 


VOLUMES. 


4 


Tetrahedron 


1.73205 


0.11785 


6 


Hexahedron 


6.00000 


100000 


8 


Octahedron 


3.46410 


0.47140 


12 


Dodecahedron 


20.64578 


7.66312 


20 


Icosahedron 


8.66025 


2.18169 



XVII. PRISMATO1D. 

1. A. IPrismatoid is a polyhedron whose bases are any two 
polygons in parallel planes, and whose lateral faces are triangles 
determined by so joining the vertices of these bases, that each 
lateral edge, with the preceding, forms a triangle with one side 
of either base 



MENSURATION. 295 

2. A Prismoid is a prisrnatoid whose bases have the same 
number of sides, and every corresponding pair parallel. 

Prob. CL. To find the volume of any prism at old. 

Formula (). ^=i(^ 1 +3^4| a )=ia(^ 2 +3^ / 2 a ), where 
a is the altitude, B ^ the area of the lower base, A^ a the area of a 
section distant from the lower base two-thirds the altitude, J3 Z 
the area of the upper base, and A\ a the area of a section distant 
two-thirds the altitude from the upper base. 

Remark. This simplest Prismoidal Formula is due to Prof. 
George B. Halsted, A. M., Ph. D., Professor of Mathematics in 
the University of Texas, Austin, Texas, who was the first to 
demonstrate this important truth. The formula universally ap- 
plies to all prisms and cylinders; also to all solids uniformly 
twisted, e. g. the square screw; also to the paraboloid, the right 
circular cone, the frustum of a paraboloid, the hyperboloid of one 
nappe, the sphere, prolate spheroid, oblate spheroid, frustum of 
a right cone, or of a sphere, spheroid, or the elliptic paraboloid, 
the groin, hyperboloid, or their frustums. For a complete 
demonstration of the Prismoidal Formula, see Halsted 's Elements 
of Geometry or Halsted' s Mensuration. 

Rule. (a) Multiply one- fourth its altitude by the sum of one 
base and three times a section distant from that base two-thirds 
the altitude. 

Formula (6). V=\a(B^M^-B^, where a is the alti- 
tude, BI and J3 2 the areas of the lower and upper bases respect- 
ively, and Af the area of a section midway between the two 
bases. 



mid 



Rule. () Add the area of the tivo bases and four times the 
d cross-section; multiply this sum by one-sixth the altitude. 



XVIII. CYLINDRIC ICINGS. 



1. A Cylindric Hing is a solid generated by a circle 
lying wholly on the same side of a line in 
its own plane and revolving abont that line. 
Thus, if a circle whose center is O be re- 
volved about DC as an axis, it will gener- 
ate a cylindric ring whose diameter is AB 
and inner diameter 2 BC. OC will be the 
radius of the path of the center O. 

FIG. 60. 
Prob. CLJ. To find the area of the surface of a solid ring 1 . 

Formula. 6'=27rrX27r7?=47r 2 r7?, whare r is the ra- 
dius of the ring, and R is the distance from the center of the ring 
to the center of the inclosed space. 





296 FINKEL'S SOLUTION BOOK. 

Rule. Multiply the generating circumference by the path of 
its center. Or, to the thickness of the ring add the inner diame- 
ter and this sum being multiplied by the thickness, and the pro- 
duct again by 9.8697044 will give the area of the surface. 

I. What is the area of the surface of a ring whose diameter 
is 3 inches and the inner diameter 12 
inches. 

By formula, 6'=4^V/?=47r 2 Xl|X 
(Iff 6)=*r 2 X 45=9.8696044X45 
444.132198 sq, in. 

1. l-in.=-J- of 3 in.=the radius r 

of the ring. 

2. 6 in.=^ of 12 in.=the radius of 

the inclosed space. 

3. 6 in.-|-l-J- in.=7-J- in. = the ra- 
il J dius R of the center of the FIG, 61. 

ring. 

4. nA C=7r3=the circumference of a section. 

5. 7r7Jr==27T/<9=27r7|=7T 15=the path of the center. 

6. .-. 7r3Xtfl5=7r 2 45=444.132198 sq. in.=the area of the 

surface of the ring. 
III. .-. The area of the surface of the ring is 444.132198 sq. in. 

Prob. CLJI. To find the volume of a cylindric ring. 

Formula. =7r V 2 /?=?rr 2 y^nR, where r is the ra- 
dius AI of the ring, and R the distance from the center of the 
ring to the center of the inclosed space. 

Rule. Multiply the area of the generating circle by the path 
of its center. Or, to the thickness of the ring add the inner di- 
ameter, and this sum being multiplied by the square of half the 
thickness, and the product again by 9. 8696044 wifl give the 
volume. 

I. What is the volume of an anchorring whose inner diame- 
ter is 8 inches, and thickness in metal 3 inches? 

By formula, V=7r*r*R=7t* X(H) 2 X(3+8)=24.75X 
9.8696044=244.2727089 cu. in. 

1. 1-| in.=J of 3 in.=the radius of the ring. 

2. 8 in.=the inner diameter. 

3. 4 in.-|-l|- in.=5|- in.=the radius R of the path of its 
,- i center. 

4. 7r(l-|-) 2 =the area of the generating circle. 

5. 27r(5|)=7T xll=the path of its center. 

6. .-. 7rllX7r(H) 2 =7r 2 X 24.75=9.86044X24.75 

=244.2727089 cu. in., the volume of the ring, 
'III, /. The volume of the ring is 244.2727089 cu. in. 



MENSURATION. 297 

THEOREM OF PAPPUS. 

If a plane curve lies wholly on one side of a line in its own 
plane, and revolving about that line as an axis, it generates 
thereby a surface of revolution, the area of which is equal to the 
product of the length of the revolving line into the path of its 
center of mass ; and a solid the volume of which is equal to the 
revolving area into the length of the path described by its center 
of mass. 

XIX. MISCELLANEOUS MEASURE- 
MENTS. 

1. MASONS' AND BRICKLAYERS' WORK. 

Masons' WOrfc is sometimes measured by the cubic foot r 
and sometimes by the perch. A perch is 16-^ ft. long, 1-J- ft. wide, 
1 ft. deep, and contains 16xHXl=24f cu. ft. 

Prob. CLJII. To find the number of perch in a piece of 
masonry. 

Rule. Find the solidity of the wall in cubic feet by the rules 
given for the mensuration of solids, and divide the product by 2^. 

I. What is the cost of laying a wall 20 feet long, 7 ft 9 in. 
high, and 2 feet thick, at 75 cts. a perch. 

1. 20 ft.=the length of the wall, 

2. 7 ft. 9 in.=7| ft.=the height of the wall, and 

3. 2 ft.=the thickness. 



II. 



4. .-. 20X71X2310 cu. ft.=the solidity of the wall. 



5. 24f cu. ft=l perch. 

6. 310 cu. ft.=310-5-24i==12H perches. 

7. 75 cts.=the cost of laying 1 perch. 

.'. 12ff X75cts.=$9.39if=the cost of laying 12ff perches. 
III. .-. It will cost $9.39^1 to Iayl2ff perches at 75 cts. a 
perch. 

2. GUAGING. 

Gauging is finding the contents of a vessel, in bushels, 
gallons, or barrels. 

Prob. CLJV. To gauge any vessel. 

Rule. Find its solidity in cubic feet by rules already given; 
this multiplied by 1728-^-215042 or .83, will give the contents in 
bushels; by 1728-^-231. will give it in wine gallons, which divided 
by Sl\ will give the contents in barrels. 

Prob. CL.V. To find the contents in gallons of a cask or 
barrel. 

Rule. (1) When the staves are straight from the bung to 
each end; consider the cask two equal frustums of equal cones^ 
and find its contents by the rule of Proh. XCIII. 



298 FINKEL'S SOLUTION BOOK 

(2). When the staves are curved; Add to the head diameter 
{inside} two-tenths of the difference between the head and bung 
diameter; but if the staves are only slightly curved, add six- 
tenths of this difference; this gives tiie mean diameter; express 
it in inches, square it, multiply it by the length in inches, and this 
product by .0084 ; the product will be the contents in wine gallons. 

3. LUMBER MEASURE. 

Prob. CL.VI. To find the amount of square-edged inch 
boards that can be sawed from a round log. 



Doyle's J^tile. From the diameter in inches subtract 
four; the square of the remainder will be the number of square 
feet of inch boards yielded by a log 16 feet long. 

I. How much square-edged inch lumber can be cut from a 
log 32 in. in diameter, and 12 feet long? 

1. 32 in.=the diameter of the log. 

2. 12 ft.=the length. 

3. 32 in. 4 in.=28 in.=the diameter less 4. 



II. 



4. 844 ft.=28 2 =the square of the diameter less 4, which 



by the rule, is the number of feet in a log 16 ft. long. 

5. 12 ft.=f of 16 ft. 

6. .'. | of 844 ft.=633 ft=the number of feet of square- 

edged inch lumber that can be cut from the log. 

III. .'. The number of square-edged inch lumber that can be 
cut from a round log 32 inches in diameter and 12 ft. long is 
633 ft. 

4. GRAIN AND HAY. 

Prob. CI/VII. To find the quantity of grain in a wagon 
bed or in a bin. 

Rule. Multiply the contents in cubic feet by 1728-^-2150.42, 
or .83. 

I. How many bushels of shelled corn in a bin 40 feet long, 
16 feet wide and 10 feet high ? 

1. 40ft.=the length of the bin. 

2. 16 ft.=the width of the bin, and 



II. 



3. 10 ft.=the height of the bin. 



4. .'. 40X16X106400 cu. ft.=the contents of the bin in 

cu. ft.. 

15. .-. 6400 X -83 bu.=5312 bu.=the contents of the bin in bu. 
III. .-. The bin will hold 5312 bu. of shelled corn. 

Rule. (1) For torn on the cob, deduct one-half for cob. 
(2) For corn not "shucked" deduct two-thirds for cob and 
shuck. 



II. 



I. How many bushels of corn on the cob will a wagon bed 
hold that is 10J feet long, 3^ feet wide, and 2 feet deep? 

1. 10i ft.=the length of the wagon bed,' 

2. 3 ft.=its width, and 

3. 2"ft.=its depth. [in cu. ft 

4. .'. 10iX3|X2=73^cu. ft.=contents of the wagon bed 

5. ,'. 73^ X-8 bu=58.8 bu.=n umber of bushels of shelled 

corn the bed will hold. 

6. .'. of 58.8 bu=29.4 bu.=the number of bushels of 

corn on the cob that it will hold. 

III. /. The wagon bed will hold 29.4 bu. of corn on the cob. 

Prob. CLVIII. To find the quantity of hay in a stack,rick, 
or mow. 

Rule. Divide the cubical contents in feet by 550 for clover or 
by 4^0 for timothy; the quotient will be the number of tons. 

Prob. CLXIX. To find the volume of any irreg-ular solid. 

Rule. Immerse the solid in a vessel of water and determine 
the quantity of water displaced. 

I A being curious to know the solid contents of a brush 
pile, put the brush into a vat 16 feet long, 10 feet wide, and 
8 feet deep and containing 5 feet of water. He found, after 
putting in the brush, that the water rose 1-J- feet ; what was the 
contents of the brush pile? 

1. 16 ft.=the length of the vat, 
TT J2. 10 ft.=the width, and 

><! 3. 1| ft.=the depth to which the water rose. 

4. .''. 16X10X11=240 cu. ft.=the volume of the brush pile. 
III. .'. 240 cu. ft=the volume of the brush pile. 

XX. SOLUTIONS OK MISCELLANEOUS 
PROBLEMS. 

Prob. CLX. To find at what distance from either end, a 
trapezoid must be cut in two to have equal areas, the divid- 
ing- line being 1 parallel to the parallel sides. 



Formula. ^=^^ 

where A is the area of the trapezoid, b the 



lower base, and b l , the upper base. \ ^(b z -\-b\) is the length of 
the dividing line. 

Rule. 1- Extract the square root of half the sum of the 
squares of the parallel sides and the result will be the length of 
the dividing line. 



300 FINKEL'S SOLUTION BOOK. 

2. Divide half the area of the whole trapezoid by half the sum- 
of the dividing line and either end, and the quotient will be the' 
distance of the dividing line from that end. 

I. I have an inch board 5 feet long, 17 inches vvidc % at one 
end and 7 inches at the other; how iar from the 
large end must it be cut straight across so that 
the two parts shall be equal? 

By formula, </=- i 



== ^(17_|_7)60-;-[V^(17 2 +7 2 )+17]=720-:-30 
=24 in.=2 ft. 

( 1. Let AB CD be the board, [end, 

2. AB=ll in.=fl, the width of the large 

3. Z>C=7 in.=y, the width of the small 

end, and [board. 

4. HK=5 ft.=60 in.==0, the length of the 

5. Produce HK, AD, and BC till they _ ___ 

meeting. Then by similar triangles, *' p lG $2. 

8. IE GL=&ED C+AB CDED C+ED C+AB C D 



II. 




9. .'. EGL=%(EDC+EAB), i. e., EGL is an arithme- 
tic mean between EAB and EDC. 

10. .-. G ! Z 2 =:^(^4^ 2 +Z>C 2 )=|(^ 2 +^ /2 ) an arithmetic 

mean between EAB and EDC, 

11. G : Z=' 



12. Draw CM perpendicular to AB. 



13. EL= %GL= 

14. IL=FLFI 

15. CM=HK=a. 



16. MB^^b'b'}. Then in the similar triangles CMB 

and CIL, _ 

17. MB\IL\\CM\CI, 01- |(^-^ 

C7. Whence 

18. C7= 



60^ v ' 'J ' '=36 in. 



177 

=3 ft. 
19. .-. 7M=CMCf=5 ft. 3 ft.==2 ft,,, the distance from 

the large end at which the board must be cut in two 
IS to have equal areas. 

III. .-. The board must be cut in two,, at a distance of 2 feet 
from the large end, to have equal areas in both parts. 

(R. H^ A^f. 4P7, greb. 101.), 



MENSURATION. 



301 



Prob. CLiXI. To divide a trapezoid into n equal parts ami 
find the leiig'th of each part. 



Formula. &!== 




the width of the small end, b the width of the large end, and a 
the length of the trapezoid. h l is the length of the first part at 
the small end, h z the length of the second part, and so on. 

I. Aboard AB CD whose length BC is 36 inches, width 
AB 8 inches and DC 4 inches, is divided into three equal pieces. 
Find the length of each piece. 

(n l) 



a 
By formula, 1= 

1)4 2 +8 2 4]=9[V32 4]=36(V2 1) =14.911686 in. 



:36[2 

1 



II. 



V3]=9.6462 in. 
4in.=the width Z>C of the small 

end, 
8 in.=the width ^4^ of the large end, 

and 

36 in.=the length BC of the board. 
.;. 216 sq. m.=%(A+DC)xC 

=i(8+4) X36=the area of the 

board. 
^ of 216 sq. in.=72 sq. in.=the area of 

each piece. 
AK=ABKB(=DC)=% in. 4 in. 

=4 in. In the similar triangles 

AKD and DCE, FIG 63 , 

AK\DK\\AB\BE, or 4 in.:36 in.::8 \n.\BE. Whence, 
^^=(36x8)H-4=72in. [triangle ABE. 




ABEABCD=2$& sq. in. 216 sq- in.=72 sq. in. 
=area of the triangle DCE. 



302 FINKEL'S SOLUTION BOOK. 

11. DCE+DCGF^Z sq. in.+72 sq. in. = 144 sq. in. 

=the area of the triangle FGE. 

12. DEC+DCGF-\-FGIH=^ sq. in.+72 sq. in.+72 

sq. in.=216 sq. in. the area of the triangle HIE. 

13. FEG\DEC\\EG*\EC*,<x 

144 sq. in.:72sq. in.:: G^ 2 :36 2 . Whence, 

14. G-fi'=^(144x36 2 )-*-72==36V2=50.911686 inches. 

15. /. GC=GE C^=50.911686 in. 36 in.=14.911686 

in. , the length of FG CD. Again, 

16. DEC:HIE\:EC*:El*,or 

72 sq. in.:216sq. in.::36 2 :^/ 2 . Whence, 

17. .57==V(216x36 2 )-r-72== 36X^3=62.3538 in._ 

18. .-. GI=EI ^=36^3 36V2=36(A/3 \/2) 

=11.442114 in., the length ofHIGF, and 

19. I=EB=EI=Tb 36V3=36(2-V3)=9.6462 in., the 

length tfABIH. 
^7=9.6462 in., 
III. ,-J G/=11.442114 in., and 
lGC=14.911686m. 

Prob. CLXII. To find the edge of the largest cube that 
c an be cut from a sphere. 



^^j ==iV 3Z>=.57735 X A where D 
is the diameter of the sphere. 

Rule. Divide the square of the diameter of the sphere by 
three and extract the square root of the quotient; or, multiply the 
diameter by .57785. 

I. What is the edge of the largest cube that can be cut from a 
sphere 6 inches in diameter? 



By formula, e=^ =1y=6x =% X 6^.57735X6 

=3.4641 in. 

{1. 6 in.=the diameter of the sphere. 
2. .'. .57735X6 in.=3.4641 in.=the edge of the largest cube 
that can be cut from the sphere. 

III. .'. The edge of the largest cube that can be cut from a 
sphere whose diameter is 6 inches, is 3.4641 in. 

Prob. CLXIII. To find the edge of the largest cube that 
can be cut from a hemisphere. 



Formula. *=-=\6X.> 408248 X^>. 



* Divide the square qf the diameter by 6, and extract the 
square root of the quotient ; or, multiply the diameter by .Jf.08248* 



MENSURATION. 303 

I. What is the edge of the largest cube that can be cut from a 
hemisphere, the diameter of whose base is 12 inches? 



By formula, e=V Z> 2 -^-6=V^=12V %=$V 6Xl2=.408248 
X 12=4.899176 in. 

yj f 1. 12 in. the diameter of the base of the hemisphere. 
|E. .-. .408248 X 12 in.=4.899176 in. 

III. .-. The edge of the largest cube that can be cut from a 
hemisphere, the diameter of whose base is 12 feet, is 4. 899176 in. 

Prob. CL.XIV. To find the diameter or radius of the three 
largest equal circles that can be inscribed in a circle of a 
given diameter or radius. 

Formula. d=D+( 1+f A/3 )=Z>^-2.1557= .4641 X# 

Kule. Divide the diameter or radius of the given cirele by 
2.1557 and the quotient will be the diameter or radius of the three 
largest equal circles inscribed in it; or, multiply the diameter or 
radius by .464!, an d the result will be the diameter or radius re- 
spectively of the required circles. 

I. A circular lot 15 rods in diameter is to have three circular 
grass beds just touching each other and the larger boundary ; 
what must be the distance between their centers, and how much 
ground is left in the triangular space about the center? 

By formula, 2r=2^?-7-(l+|V / 3)==2^?-r-2.1557=^. T 1 3 Vr 
=6.9615242 rd.=the distance between their centers. 

Construction. Let AHE be the circular lot, C the center, and 
A CE any diameter. With E as a center and radius equal to 
CE describe an arc intersecting the circumference of the lot in H. 
Draw a tangent to the lot at E and produce the radius CH to 
intersect the tangent at B. Bisect the angle CBE and draw 
the bisector GB. It will meet the radius CE in G, the center 
of one of the grass beds. Draw GF perpendicular to CB. Then 
GF=GE, the radius of one of the grass beds. Draw EH. 
Then EH=CH~EC, and CH=HB, because the triangle 
EHB is isosceles. 

1. CZ=7-| rd.==#, the radius of the lot. 

3. EB^VCBICE^VCIR^R^RVZ. In the 

similar triangles CFG and CBE, 

4. CF\FG\\CE\\EB,w CF:GF:\R\RV%. But 

5. CF= 

^6. .'. 7?(2 VZ)\GF\:R'M^%- Whence, 



.4641=3.48075 rd.=the radius. 



304 



FINKEL'S SOLUTION BOOK. 



11. 



=2r=2X(23 3)=6.9615rd., the distance be- 
tween their centers. 



2. 



3. 




area of the triangle 
IGK. 
Area DKF=\ of the 

small circle, because 
the angle' DKF is 
60, or \ of 360. 
. .'. Area 

5. -J7rr 2 := 

=the area of the FIG. 64 

three parts of the small circles within the triangle 

IGK. 

6. .-. r 2 A/3 i7rr 2 =^ 2 (V / 3 |7r)=.16125368r 2 

.16125368X[^(2V3 3)] 2 =.16125368X('21 

12A/3)./? 2 =:.16125368x.2153904x^ 2 
=.03473265 X^ 2 =-03473265x(7^) 2 =1.953712 
sq. rd.=the area of the space inclosed. 

{6.9615 rd.=the distance between their centers, and 
1.953712 sq. rd.=the area inclosed about the center of 
the given lot. (R. H.A., p. 4.07, prob. 100.) 

Prob. CL.XV. Having- given the area inclosed by three 
equal circles to find the radius of a circle that will just in- 
close the three equal circles- 






~Nv 03473265>)' 



where A is the area inclosed - 



Rvile. Divide the area inclosed "by .034.73265 and extract the 
square root of the quotient, and the result will be the radius of the. 
required circle. 

Prob. CLXVI. Having given the radius a, b, c, of the 
three circles tangent to each other, to find the radius of a 
circle tangent to the three circles. 

abc 

Formula. r or r'= n . . r-j ::- - j-r^r> 

1\/[abc(a+b+c)] =F ( ab-\-ac-\-bc) 

the minus sign giving the radius of a tangent circle circumscrib- 
ing the three given circles and the plus sign giving the radius 
of a tangent circle inclosed by the three given circles. 

NOTE. This formula is due to Prof. E. B. Seitz, Late Professor of Mathe- 
matics in the North Missouri State Normal School, Kirksville , Mo., of 
whom we give a biographical sketch accompanied by his photograph. 

This for mula is taken from the School Visitor, Vol. II. p. Ill, with the 



MENSURATION. 305 

slight change that the plus sign is introduced for the case in which the 
tangent circle is inclosed by the three given circles. The problem of finding 
two circles tangent to three mutually tangent circles, is one supposed 
to have been proposed by Archimedes more than 2000 years ago, though 
the problem he proposed was not so general the diameter of one of the 
given circles being equal to the sum of the diameters of the other two. 

The problem of finding all circles that can be drawn within three mu- 
tually tangent circles and tangent to each of them, has been simply and 
elegantly solved by D. H. Davison, Minonk, 111. The above formula led 
him to his wonderful solution. For a complete and elegant solution, 
where he has actually computed and constructed 81 circles tangent to three 
given circles, see School Visitor, Vol. VI , p. 80. 

Prob. CLXVII. To find the surface common to two equal 
circular cylinders whose axes intersect at right angles. 

Formula. S16jR 2 , where R is the radius of the cylinders. 

Rule. Multiply the square of the radius of the intersecting" 
cylinders by J6. 

I. If the radius of two equal circular cylinders, intersecting 
at right angles is 4 feet, what is the surface common to both? 

By formula, S= 16^? 2 = 16 X4 2 =256 sq. ft. 

f 1. 4 ft. = the radius of the cylinders. 

II. J 2. 16 sq. ft. = 4 2 = the square of the radius of the cylinders 
) 3. .-. 16 X 16 sq. ft. = 256 sq. ft. = the surface common to the 

two cylinders. 

III. .. 256 sq.ft. = the surface common to the two cylinders. 

Prob. CLXVIII. To find the volume common to two equal 
circular cylinders whose axes intersect at right angles. 



Formula. V= V^ 8 wnere R * s tne radius of the 
cylinder. 

Rule. Multiply the cube of th0 radius of the cylinders 
ly V . 

1. A man digging a well 3 feet in diameter, came to a log 3 
feet in diameter lying directly across the entire well; what was 
the volume of the part of the log removed ? 

By formula, V= \fR* = V(l) 8 = 18 cu - ft 

1. 3 ft. = the diameter of the log and the well. 

2. 11 ft. = the radius. 



II 



3. 3| cu. ft. = ( 1 1) 3 = the cube of the radius. 



4. .. V 6 X 3f cu. ft.= IScu.ft. , the volume of the part of 

the log removed. 
III. .'. The volume of the part of the log removed is 18 cu.ft. 

Prob. CLXIX. To find the height of an object on the 
earth's surface by knowing its distance, the top of the ob- 
ject being visible above the horizon. 



306 FINKEL'S SOLUTION BOOK. 

Let J3F=a be any object, AJ3=t a tangent to the earth's sur- 
face from the top of the object, and FE-=D the diameter of the 
earth. Then by Geometry, AB*=BF(BF+FE),m t' 1 =a(a 

t* 
-{-/)). .*. a= t ^-. But a is very small as compared with the 

diameter of the earth and AB=AF without appreciable error. 

AF Z c* 
.'. Formula. a = =-rr, where c is the distance to 

the object from the point of observation. 

I 2 
When c=\ mile, a= =-| ft., nearly. 



Rule. Multiply the square of the distance in 
miles by\, and the result 'will be the height of the 
object in feet 

I. What is the height of a steeple whose top can 
be seen at a distance of 10 miles? 

c 2 10 2 10 2 
By formula, === 




{1. 10 miles=the distance to the steeple. 
2. 100=10 2 =the square of the distance. 
3. .-. f of 100= 66| ft.=the height of th 



III. .-. The height of the steeple is 66f ft. 

Prob. CLXX. To find the distance to an object by kno\v- 
ing its height, the top only of the object being visible above 
the horizon. 



Rule. Multiply the height of the object in feet ~by | and ex- 
tract the square root of the product , and the result 'will be the dis- 
tance in miles. 

I. At what distance at sea can Mt. Aconcagua be seen, if its- 
height is known to be 24000 feet? 

By formula, c=Vj^=V% X24000=V36000=190 mi., nearly. 

il. 24000 ft.=the height of the mountain 
2. 1X24000=36000. 
3. .-. V 36000=10 V360=190 mi., nearly. 
III. /. Mt. Aconcagua can be seen at a distance of 190 miles. 

Prob. CXXXI. Given the sum of the hypotenuse and 
perpendicular, and the base, to find the perpendicular. 

S 2 _ ^2 

Formula. /== - - , where s is the sum of the hy- 
potenuse and perpendicular, and b the base. 



MENSURATION. 307 

Rule. 1- From the square of the sum of the hypotenuse 
and perpendicular subtract the square of the base, and divide the 
difference by twice the sum of the hypotenuse and perpendicular. 

2. To find the hypotcmise: To the square of the sum of the 
hypotenuse and perpendicular, add the square of the base and di- 
vide this sum by twice the sum of the hypotenuse and perpendic- 
ular. 

I. A tree 120 feet high is broken off but not severed. The 
top strikes the ground 34 feet from the foot of the tree; what is 
the height of the stump? 

5 2_ 2 120 2 34 2 [stump. 

By formula, /= ^ :___=55J ft., the height of the 

1. 120 ft.=the sum of the hypotenuse and perpendicular. 

2. 34 ft.=the base, or the distance the top strikes from the 

foot of th~ tree. 



II. 



3. 14400 sq. ft.=120 2 =the square of said sum, 



4. 1156 sq. ft.=34 2 =the square of the base, and 

5. 14400 sq. ft 1156 sq. ft.=13244 sq. ft.=the difference. 

6. .-. 13244-r-(2Xl20)=55^ ft.=the height of the stump. 

III. .-. The height of the stump is 55^ feet. 

NOTE. This rule is easily derived from an algebraic solution. Thus: 
Let #=the perpendicular, s # the hypotenuse, and fcrthe base. Then, 



x*-\-b*=(s *) 2 , or x z-\-l>2=s* 2sx+x*, and y ~. 

Prob. CLXXII. To find at what <listance from the large 
end of the frustum of a right pyramid,a plane must he passed 
parallel to the hase so that the two parts shall have 
equal solidities. 

3 V 
Formula. h=- - - , where V is the 



volume of the frustum, B the area of the lower base, B 2 the area 
of the "dividing base," andVj2J5 2 the area of the mean base be- 
tween the "dividing base" and and lower base. 

Rule. 1. Find the -volume of the frustum by Prob. XCIII. 

2. find the dimensions of the "dividing base" by extracting the 
cube root of half the sum of the cubes of the homologous dimensions 
of the upper and lower bases. Then find the area of the ^divid- 
ing base" 

3. Divide half the volume of the frustum by one-third of the 
sum of the areas of the lower base, "dividing base," and mean 
base between them, and the quotient will be the length of the 
lower part. 

I. How far from the large end must a stick of timber, 20 feet 
long, 5 inches square at one end and 10 inches square at the 
other, be sawed in two parts, to have equal solidities? 



FINKEL'S SOLUTION BOOK. 



By formula, /i=j - 



240(10 2 +10x5+5 2 ) 



42000 



1680 



2(100+25^36+^2-^6) 8+2^36+^6 
1680 1680 



8+6.603855+5.4513618 20.0552168 



7e 3 . 



h$ piece of limber, ABCl> 
the lower base, EFGH the upper base, and OL the altitude. 
Prolong the edges AH, BE, CF, and DG and the altitude <9Z 
till they meet in P. Draw KL to the middle point of AD, Ol 
to the middle point of GH and draw PIK. Let SMNR be 
the dividing base. 

1. ^4^=10 iti.=, the side of the lower 

base. 

2. HE=Z> in.=<:, the side of the upper 

base, and 

3. OZ=20 ft.=240 in.=, the altitude. 

4. KQ=KL QL(=IO)=\(b c)=i(10 

in. 5 in)=2-Jin. By similar triangles, 

5. ^^: QI::KL:PL, or 

Whence, 

ft. 



II. 



/. 

8. 
9. 

10. 
11. 
12. 



b <: 



c 



=20 ft. 




FIG 66 
=2000 cu. in., the volume of the pyramid HEFG. 



13. 
14. 



+c 2 )=14000 cu. in., the volume of the frustum 

ABCDE. 
.-. F=of 14000 cu. in.=7000cu. in., the volume of 

each part. 
V+-J F==2000 cu, in.+7000 cu. in.=9000 cu. in., the 

volume of the pyramid, SMNR P, and 
v+ K=2000 cu. in.+14000 cu. in.=16000 cu. in., the 

volume of the pyramid ABCD P. By the princi- 

ple of similar solids, \AB*, or 

HEFG-P : SMNP-P : ABCD-P : : HE* : SM* : 
v : v V-.V V. : c 3 : 5J/ 3 : b* . But 



MENSURATION 309 

15. ,4-1 V = =fi v +(v+ F)], i. e., *-H V, or SMNRP\s an 
arithmetical mean between v and *;-J- Fj or HEPG 



17. .-. SM*=(c ? +&*),{. e, $/J/ 3 is an arithmetical mean 

between ./7 3 and AB r >, or c 3 and 3 . Whence, 

18. 5^=== 

8.2548188-Mn. 

19. SJ/ 2 =y[| (^-f 3 ) ] 2 =( fy 36 ) 2 =Y' ^6=68. 14202 

sq. inr=the area of the dividing base. 
20 V(S^X^^ 2 )=SJ/x^#==if / 36xlO=25y36== 

82. 54818 sq. in.=the area of the mean base of the 

part cut from the frustum. 
21. /. 



+y (36) XlO+(fy36) 2 ]=i:T(100-f 82.54818 
+68.14202)= ^ZT^X 250.6902 =-Zrx 83.5634 = the 
volume of the fru&tumACDM .But [ M. 

23. | F=7000 cu. in.==the volume of the frustum ABCD 

24. .'. LTX 83.5634=7000 cu. in. Whence, 

25. Zr==7000-^83.5634==83.76883 in.==6 ft. 11.76883 in., 

the length. 

III. .'. The stick must be cut in two at a distance of 83.76883 
: in., or 6 ft. 11.76883 in., from the large end. 

NOTE. The frustum of a cone may be divided into two equal parts in 
the same manner. The frustum of a pyramid or a cone can be divided into 
any number of equal parts on the same principle as that for dividing a 
trapezoid into n equal parts, Prob. CLXI. 

I. The area of a rectangle whose length is 20 rods is 120 sq. 
rods; what is the area of a similar rectangle whose length is 30 
xods? 



>, Similar areas are to each other as the squares 
of their like diinentions or as the squares of any other homologous 
dines. 

'1. 20 rods=the length of the given rectangle, and 

2. 120 sq. rd.=its area. 

3. 30 rods=the length of the required rectangle. 
J4. .'. 20 2 :30 2 ::120sq. rd. : (?). Whence, 

[5. ?= (120x30 2 )-r-20 2 =270 sq. rd. 
III. .. The area of the rectangle is 270 sq. rd. 

I. The area of a rectangle whose width is 7 feet, is 210 sq. ft. ; 
what is the length of a similar rectangle whose area is 2100 sq. ft. 

1. 210 sq. ft =the area of given rectangle, and 

2. 7 ft.=its width. Then 
TT 3. 210-7-7=30 ft its length. 

' 4. /. 210 sq. ft: 2100 sq. ft. : : 30 2 :( ?). Whence, 
5. ?=(2100x30 2 )-:-210=300 ft.=the length of the re- 
quired rectangle. 



II, 



310 FINKEL'S SOLUTION BOOK 

III. The length of the required rectangle is 300 feet. 

I. If the weight of a well proportioned man, 5 feet in height, 
be 125 Ibs., what will be the weight of a similarly proportioned 
man 6 feet high? 

Principle. Similar solids are to each other as the cubes of 
their like dimensions or as the tubes of any other homologous lines. 

1. 5 ft.=the height of the first man, and 

2. 125 lbs.=his weight. 

3. 6 ft the height of the second man. 

4. /. 5 3 :6 3 ::1251bs:(?). Wivnce, 

5. ?=(125X6 3 H-5 3 =216 Ibs., the weight of the second 

man. 
III. .-.The weight of the man whose height is 6 feet, is 216 Ibs. 

I. James Page has a circular garden 10 rods in diameter. How 
many trees can be set in it so that no two shall be within 10 feet 
of each other and no tree within 2-J- feet of the fence? 

Construction. LetAJB C be the circular garden, AC it diameter, 
and O its, center. Then with O as a center and radius A O = -J-of 
(10X16|- ft 2X2| ft.), or 80 ft, discribe the circle abcdef, 
and in it describe the regular hexagon abcdef. Then aO 
=tf&=80 ft. Begin at the center of the circle and put 
8 trees 10ft. apart on each radii, aO, 60, cO, dO. eO, and/0. 
Then joining these points by lines drawn parallel to the diame- 
ter of the circle as shown in the 
figure, their points of intersec- 
tion will mark the position of the 
trees. Hence, the trees are ar- 
ranged in hexagonal form about 
the center. The first hexagonal 
row contains 6 trees, the second, 
12, the third 18, and so on. Since 
the radius of the circle on which 
the trees are placed is 80 feet and 
the trees 10 feet apart, there will 

be 8 hexagonal rows. 

FIG. 67, 

1. 6=the number of trees in the first hexagonal row. 

2. 12=the number of trees in the second hexagonal row. 

3. 48=the number of trees in the eighth hexagonal row. 

4. .. 216=4(6+48)X8=the number of trees in the eight 




II, 



hexagonal rows. 

5. 24=6x4=the number of trees at the sides of the hexa- 

gon abcdef. 

6. .'. 216+24+1, the tree at the center ,=241=the number 

of trees that can be set in the garden. 



MENSURATION. 



311 



III. /. There can be set in the garden, 241 trees. 

( Greenleafs Nafl Arith., />. 444, prob. 25.} 

I. There is a ball 12 feet in diameter on top of a pole 60 feet 
high. On the ball stands a man whose eye is 6 feet above the 
ball; how much ground beneath the ball is invisible to him? 

Construction. Let BE be the pole, L the center of ball, and 
A the position of the man's eye. Draw AFC tangent to the ball 
atT? and draw LFwdBC. Then the trbnglc AFL is right- 
angled at F. 

'1. 60 it.=BE, the length of the 
pole. 

2. 12 ft.=J?Z>, the diameter of 

the ball, and 

3. 6 ft.=A>, the height of the 

man's eye above the ball. 
12 ft. AD+DL=AL. Now 



II. 



6. 



7. 



8. 



=V(12 2 6 2 )=6\/3ft. In the 

similar triangles ALF and 
ACB, 
AF-.LF: :AB: BC, or 

6V3 ft. :6ft. ::(6ft.+12ft. 

+60 ft.), or 78 ft.: BC. 
.-. BC=( 6x78)-5-6V3=78 

-2-V3=jx78V3=26V3 ft. 




the circle over which the man can not see. 



FIG 68. 

sq. ft.=the area of 



III. .-. 6371.1498932 sq. ft.=the area of the invisible ground 
beneath the ball. 

I. Three women own a ball of yarn 4 inches .in diameter. 
How much of the diameter of the ball must each wind off, so that 
the may share equally? 

1. 4 in.=the diameter of the ball. Then 

2. i-7r(4) 3 ==\ 2 7r=the volume of the ball. 

3. ij- of 3 3 2 7r= 3 g 2 7T=each woman's share. 

4. 3 -^7T 3 9 2 7r= 6 g 4 7T=the volume of the ball after the first 

has unwound her share. But 

7fjD' d ==the volume of any sphere whose diameter is />. 
i 6 ^ 4 7r. Whence, 



II. 



6. 

8. Z>=^/ 9l |8^4^| = l4^ 18= 4 X 2.6207414=3.4943219 in., 

diameter of the ball after the first unwound her share. 

9. .'. 4 in. 3.4943219 in=.5056781 in., what the diameter 

was reduced by the first woman. 

30. --7t*g7t=*-g7t, the volume of the ball after the 
second had unwound her share. 



-312 FINKEL'S SOLUTION BOOK. 



11. .-. y(V*r-H?r)=4yi=fy 9=JX2.0800837 

=2.5734448 in., the diameter of the ball after the sec- 
ond woman unwoud her share. 

12. .-. 3.4943219 in. 2.5734448 in.=.7208771 in., what *he 

diameter was reduced by the second woman. 

The diameter was diminished .5056781 in. by the first 

woman, 

.7208771 in. by the second woman, and 
2.7734448 in. by the third woman. 

(Milne's Prac. Arith.,p. 335, prob. 8.) 

NOTE. The following are the formulas to divide a sphere into n equal 
parts, the parts being concentric: 



* nd 



is the diameter of the sphere; Z^the diameter after the first 
part is taken off; Z> 2 , the diameter after the second part is taken 
off; and soon. Then/? Z> 1? D^ Z) 2 , &c, are portions of 
the diameter taken off by each part. 

I. A park 20 rods square is surrounded by a drive which con- 
tains j 1 ^- of the whole park; what is the 
width of the drive? 

1. 20rd.=^4/>=DC, a side of the 

park. 

2. 400 sq. rd.=20 2 = the area of the 

park A BCD. 

3. T Vo- of 400 sq. rd.=76 sq. rd.=the 

-II. ' 



area of the path. 




4. 400 sq. rd. 76 sq. rd=324 sq. rd. 

=the area of the square EFGH. FIG. 69. 

5. ^^=V(324)=18 rd., the side of the square EFGH. 

6. .'. 7/T .^^=20rd. 18rd.=2rd., twice the width of the 

path. 

17. .'. 1 rd.=| of 2 rd.=the width of the path. 
III. .-. The width of the path is 1 rod. 

I. My lot contains 135 sq. rd., and the breadth is to the length 
as 3 to 5 ; what is the width of a road which shall extend from 
one corner half around the lot and occupy \ of the ground. 

Construction. Let ABCD\>z the lot, and DABSNR the 
road. Produce AB, till BE is equal to AD. Then AE is 
equal toAB-\-AD. OnA, construct the square AEFG, and 



MENSURATION. 



313, 




on EF and GF respectively, lay off El and T^VTequal to AB. 
Then construct the rectangles BEIH, ILKF, and KMDG. 
They will each be equal to A BCD, for their lengths and widths 
are equal to the length and width of ABCD. Continue the road 
around the square. Then the area of the road around the square is 
four times the area of the road DABSNR. 

1. =the width AD of the 

lot. Then 

2. f=the length AB. 

3. 1x1=135 S q. rd., the 

area of the lot. 

4. Xt=T of 135 sq. rd. 

=27 sq. rd., and 
5- fXf=(f) 2 = 3 times 27 
sq.rd.=81 sq. rd. 

6. /. i=V81= 9rd.,the 

width AD, 

7. = of 9 rd.=3 rd., and 

8. 4=5 times 3 rd.=15 

rd.,the length AB. 

9. 15 rd+9 rd.=24 rd.= FIG. 70. 

AjB,the side of the square AEFG. 

10. .-. 576 sq. rd.=24 2 = the area of the square AEFG. 

11. 33| sq. rd.=i of 135 sq. rd.=the area of the road 
DABSNR. 

12. .-. 135 sq. rd.=4x33j sq. rd.=the area of the road 
around the square. Then 

13. 576 sq. rd. 135 sq. rd.=441 sq. rd., the area of the 

square NOPQ. 

14. .-. 21rd.=V441 =NO, a side of the square NOPQ. 

15. AE NO=24; rd. 21 rd.==3rd.=twice the width of 

the road. 

16. .'. H rd.=24| ft.=-J- of 3 rd.=the width of the road. 

III. /. The width of the road is 24| ft. 

(jR. H. A.,p.407,pr0b.99.) 

I. The length and breadth of a ceiling are as 6 to 5 ; if each 
dimension were one foot longer, the area would be 304 sq. ft. ; 
what are the dimensions? 

Construction. Let ABCD be the ceiling, AB its width and 
BC its length. Let AIGE\>t the ceiling when each dimension 
is increased one foot. On BC, lay ofF^^fequal to AB and draw 
L,K parallel to AB. Then ABKL is a square whose side is. 
the width of the ceiling. 



314 



FINKEL'S SOLUTION BOOK 



II. 



1. l=AB, the width of the ceil- 

ing. 'Then 

2. \=BC, the length, and 

3. -fxf=^^X^C=the area of 

the ceiling. 

4. fXl=^CX#/, El being 1 

foot,=the area of the rect- 
angle B CHI. 

5. |Xl= DCxCF, CF being 

1 foot,=the area of the rect- 
angle DCFE. 

6. 1 sq.ft.=! 2 =the area of the 
square CFGH. 

7- .-.fXf+fXl+fXl+lsq.ft. 
=the area of AIGE. But 



9. . 

10. 
11. 
12. 
13. 



DCFE, 

f+f , <> r 

fxf+fxi+- 5 




14. 
15. 

16. 

17. 

18. 

19. 



20. 
21. 

22. 



23. 
24. 



i.e., it equals a rectangle whose length is 
TT 1 , and width 1 ft. 

IXl+lsq. ft.=fXf+y Xl+1 sq. ft. 
=the area of AIGE. But 
304 sq. ft.=the area of AIGE. 
.' f Xf+y Xl+l sq. ft.=304sq. ft. 
I Xf+y X 1=303 sq. ft=the area of 
y=y Xf in which y is y ft. ; for a rectangle whose 
length is y , and the width 1 ft, has the same area as 
a rectangle whose width is y ft. and length 1, orf. 
' f Xf +y Xf=303 sq. ft, in which y is y ft 

ixf+Hxf=50i sq. ft.=j. of (fxf+yxf)=i of 

303 sq. ft., 



Whence, 
AIHCFE.Kut 



sq. ft. But 
|xf=the area of the square ABKL, and 
y Xf the area of the rectangle ALNP whose length 

AL is f and width LN y ft. 
iof(Xt)=HXf=haIf the rectangle ALNP=the 

rectangle OMNP, which put to the side AB of the 

square ABKL as in the figure. 

|Xf+y Xf=252^ sq. ft.=thearea of SRAOMK. 
\\\ sq. ft.=(J-J) 2 ==theareaof the square RQOA, since 

AR is \\ ft. 
...fxf+yXf+Hi sq. ft.=252i sq. ft.+ifi sq. ft. 

= ^648i sq ftthe area of (SRAOMK+RQOA). 

=the area of SQMK. 

f+H ft.=V-fll i=:1 T ft.=the side 5"^ of the square 

SQMK. 
iHVV ft. Ji- ft.=V 2 ft.=15 ft.= 

=Afi t the width of the ceiling. 



MENSURATION. 



315 



TTT 



25. i=4 of 1-5 ft.= 3 ft., and 

^26. f=6 times 3 ft.=18 ft.= 7?C, the length of the ceiling. 
C 15 ft.=the width of the ceiling, and 
\ 18 ft.=the length. 
Remark. In this solution there is but one algebraic operation ; 
viz., extracting the square root of the trinomial expression, 
(|X|+y Xf+yfi s q- in -)> in ste P 23. This might have been 
omitted and then the solution would have been purely arithmeti- 
cal ; for, the area of the square SQMK being known, as shown 
by step 22, its side SK could have been found by simply extract- 
ing the square root of its area, 2 -^!- 1 - sc l- ^- Then by subtracting 
SB, which is \\ ft, from SK, we would get BK(=AB), the 
width of the ceiling. 

The following solution is quite often given in the schoolroom: 
304 -=-(5X6) = 10+. >/10 = 3+. 
5 X3 = 15, the width and 6 X 3=18, the length. 

I. A tin vessel, having a circular mouth 9 inches in diameter, 
a bottom 4-j inches in diameter, and a depth of 10 inches, is part 
full of water ; what is the diameter of a ball which can be put 
in and just be covered by the water? 

Construction. Let AB CD be a vertical section of the vessel, 
AB the top diameter, DC the bottom diameter, and EF the al- 
titude. Produce AD, BC, and EF\\\\ they meet in G. Draw 
MC parallel to EF. In the triangle A CB inscribe the largest 
circle IEP arid let Q be its center. Draw the radius IQ. Now 
( 1. AE=^AB=R=^ in.=the radius of the mouth. 

2. CF=DC=r=2 in., the radius of the bottom, and 

3. EF=a=lO in., the altitude of the vessel. 



2 in.=2J in. In the similar triangles 



5. MB-.MC-. :EB\EG,or 

Rr la-.iR-.EG. Whence, 



*=20 in., the altitude 



of the triangle A GB. 

1&AGB 
* v~ 



AB+A G+BG~AB+A G+BG 



4JRa 




==3f in., the radius 
of the largest sphere that can be put in the vessel or in 



316 FINKEL'S SOLUTION BOOK. 

the cone A GB. 

10. 

=the volume of the largest sphere that can be put: 
in the cone AGB. 

11. EGxx 2 =i7t2aR 2 =$7f X20X(4|) 2 =1357r, the 
volume of the cone A GB. 



12. ... i*^-|* ^.^ X 



the quantity of water in the cone which will just cover 
the largest ball that can be put in the cone A GB. 

13. ^7r^GX^C 2 =i^r 2 =^7rXlOX(2i) 2 =-i-f^7r, the 

volume of the cone DGC. 

14. .-. i7Ttf7- 2 +i of the volume of the vessel = if^+i of 

the volume of the vessel the quantity of water in 
the cone necessary to cover the required ball. But 

15. $7ta(R*+Rr+r* )=i7rlO[(4|) 2 +4|X2i+(2i)2] 

f 5.7T, t h e volume of the vessel, by Prob. XCIII. 

16. .*. nar 2 -- f tne volume of the vessel =rtar 2 -- of 



7T-|~|-of - 9 - 4 r -7T=^-|f-7T, the quantity necessary to cover 
the required ball. 

17. .'. The quantity of water necessary to cover the largest 
ball: the quantity of water necessary to cover the 
required ball : : (radius) 3 of largest ball : (radius) 3 of 
required ball. Hence, 



19. 9_o^9^ : i4|5^ . : (3|)3 : JfO*. Whence, 

20. y337:|^ 55 ^3 :HO. Whence, 



21. JYO= 



and 

22. 18f (^)=6.1967-hin., the diameter of the required 
ball. 

III. /. The diameter of the required ball is 6.1967+ in. 

I. I have a garden in the form of an equilateral triangle 
whose sides are 200 feet. At each corner stands a tower; the 
height of the first tower is 30 feet, the second 40 feet, and the 
third 50 feet. At what distance from the base of each tower 



MENSURATION 



317 



must a ladder be placed, so that without moving it at the base it 
may just reach the top of each, and what is the length of the 
ladder? 

Construction. Let ABC be the triangular garden and AD, 
BE, and CF the towers at the corners. Connect the tops of the 
towers by the lines ED and DF. 
From G and H, the middle points 
of DB and DF, draw GJ/and HN 
perpendicular to DE and DF, and 
at M and N draw perpendiculars to 
AB and A C in the triangle ABC, 
meeting at O. 1 lien O is equally 
distant from D and E. For, since 
M is equally distant from D and E, 
and MO perpendicular to the plane 
ABED, every point of MO is 
equally -distant frc i D and E. For 
a like reason, every point of NO is p[ G 

equally distant frcl_i D and F\ hence, O their point of intersec- 
tion, is equally dJ^ant from D, E, and F and is, therefore, the 
point where the ladder must be placed. Draw _Z?/and DJ par- 
allel to AB and A C, G^f and HL perpendicular to AB and A C, 
MP perpendicular to A C, and OR parallel to NP. Draw the 
lines OB, OC, and OA, the required distances from the base of 
the ladder to the bases of the towers. Draw EO, the length of 
the ladder. 



1. AB= C=A C=200 ft.=j, the side of the triangle. 

2. ^C=50 ft.=0, the height of the first tower, 

3. -Z/?=40 ft.=$, the height of the second tower, and 

4. AD=Zb ft.=c, the height of the third tower. Let 

5. h=*J[AB*(A C) 2 ]=V[> 2 (i*) 2 ]= iV3*=100 

Y/3 ft=the perpendicular from J? to the side A C. 

6. &f=J3f=AD=6c=40ft.3Qft. 




=10 ft. 

:jB+AD)=\(b+c)=$(4Q ft.+30 ft.) 
=35 ft. In the similar triangles DIE and GKM, 
DI-.IE : : GK\ KM, or 5 : b c : : $(b+c) : KM. 

* c 2 _4Q 2 30 2 f 

~Zs 2X200"" ~~ f '' 



7. G^= 



8. 



10. AM= 

= 101 j ft, and 



=98^ ft. In like manner, 
12. J7=i(+c)=i(50 ft.+30 ft.)=40 ft., 



318 



FINKEL'S SOLUTION BOOK. 



II. 



,2 r 2 



13. LN= ^ 4 ft, 



14. AJV= 



15. JVC=ACAN=s 



2s 2s 

+ o o Of 

gg g a j2-f ( 

2s < 



=104 ft. 



96 ft. By similar triangles, 
' 



16. 

: AP. Whence, 
17. AP= 
18- .-. 




19. J?O=PN = 

C 2 )^.2s 
similar triangles, 

20. AB'.BL: :AM:MP,ors: 

MP. Whence, 

21. AfP= 
22. 



[lar triangles, 
Bysimi- 



23. XM=(s*+2a2 ^ 2 

c 2 )-f-125]V3=17HV3 ft. Again 

24. MP-.MA '.:RO-. OM, 



or 



: OM. 



25. .-. OyT/^ 

ft 

26. O N=R P=MPR M= 



2 2 ^V 2 



-f-65] v /3=33| V3 ft- Then 

27. OC 

(9 
L 

=111.8796+ft. 

28. 0^ = 



=V14116 T L=1 18.81 11+ft. 



29. 






B. =^ 

length of the ladder. 



MENSURATION. 



319 



1. 111.8796-fft.=the distance from base of the ladder- 

to the base of the tower FC, 

2. 1 18.81 11+ft. the distance from the base of the lad- 
Ill, .'.'i <Jer to the base of the tower AD. 

'3. 115.8278+ft.=the distance from the base of the lad- 
der to the base of the tower BE, and 
4. 122.5402+ft.=the length of the ladder. 

( Greenleaf's Nat* I Arith., p. 444, prob. 38.) 
Remark. When the sides of the triangle are unequal, pro- 
ceed in the same manner as above. In some cases the base of 
the ladder will fail without the triangle. 

I. At the extremities of the diameter of a circular garden 
stands two trees, one 20 feet high and the other 30 feet high. At 
what point on the circumference must a ladder be placed so that 
without moving it at the base it will reach to the top of each tree, 
the diameter of the garden being 40 feet. 

Construction. Let ABC be the circular garden and A C its 
diameter, E nd let AF and CD be the two trees at the extremi- 
ties of the diameter. Connect the tops of the trees by the line 
FD and from the middie point E of FD let fall the the per- 
pendicular EH. Draw EG perpendicular to FD. Then all 
points of EG are equally distant from FD. At G draw BG 
perpendicular to AC. Then all points of BG are equally dis- 
tant from F and D. Hence, B is the required point. 

1. A C=2/?=40ft., the diameter of the garden. 

2. C0==30 ft., the height of the tree CD, and 

3. AF=.--b=Z(j ft, the height of the tree AF. 



II. < 



40 ft. 30 ft.=lO ft. 

5. EH--=- 1 ( CD+ AF) =4 (a+b) 

=(40 ft+30 ft.) =35 ft. By 
similar triangles, 

6. FI\ ID : : EH-. HG, or 

Whence, 

7. HG= C ^^=m ft 




FIG. 74. 




4 (* 2 <* 2 ) 2 -] 

-167? = 



ft. 



=34.91165 ft., nearly, and 



320 



FINKEL'S SOLUTION BOOK. 



10. ^c= 



82 f,=11.31942 ft. 



34.91165 ft. the distance from the smaller tree, and 
11.31942 ft. the distance from the larger tree. 

I. Seven men bought a grindstone 5 feet in diameter ; what 
part of the diameter must each grind oft' so that they may share 
equally? 

Construction. Let AB be the diameter of the grind stone, 
O its center, and A O its* radius. From A draw any indefinite 
line AW and on it lay off any convenient unit of length seven 
times, beginning at A. Let/* 
be the last point of division. 
Draw OP ', and from the other 
points of division draw lines 
parallel to OP, intersect- 
ing the radius AO, in the 
points /, e, d, c, 6, and a. Then 
the radius is divided into 
seven equal parts. On radius 
AO, as a diameter describe a 
semi^circumference A OK,and 
at a, b, c, d, e, and/", erect per- 
pendiculars intersecting the 
semi-circuro^ence in M, L, 
K, 7, H, ana G. Then with 
O as a center and radii equal the chords MO, LO, KO, 7(9,. 
HO, and GO, describe the circles as shown in the figure. Then 
each man's share will be the area lying between the circumfer- 
ences of these circles. For, the chord GO 2 =Gf 2 -\-fO 2 and, by 
a property of the circle, Gf 2 =AfxfO. .'. GO^^AfxfO+fO*, 
= (Af-\-fO)fO=:A OxfO=\A O 2 . In like manner HO*=A O 
X 0=4^4 O 2 , KO 2 =^AOxdO=^AO 2 , &c. 




1. AB=D=5 ft., the diameter of the grind stone. 

2. A O=R=1% ft., the radius. 

3. .'. rt/? 2 =7rx(2!) 2 =6^7r=the area of the stone. 

4. |of7T/? 2 :=i7r^ 2 =|of 6i7r==ff7r==each man's share. 

5. Q^Tf |f 7Tr=i|^r=the area of the stone after the first has 

ground ofT his share. 

6. .-. V(i|7r^-7r)= 1 \y'42=2.31455+ft., the radius MO. 

7. 2(AO MO)=2(2% ft T \V42 ft.) =2 (2| ft. 2.31455 

ft)=.3709 ft, part of the diameter the first grinds off. 

8. Q^TT 2 O fgj 7r== y ff 5^ === the area after the second grinds 

off his share. 



MENSURATION. 



321 



9. .-. V(W 7r -^ 7r ) = f\ / 7= =2 - 112 875 ft., the radius LO. 
Then 

10. 2(MO 0)=2(5v / T V- f\/f )=2(2.31455 ft 

2.112875 ft)=.40335 ft, the part of the diameter the. 
second grinds off. 

11. 6i?r f cf 6i7T= 2 T 5 7r=the area after the third has 

ground off his share. 

12. .-. V( 3 T 5 7r-7-7r)=|\/4=5V'y=1.889822-4- ft., the radius 
"KG. Then, 

1.8899822 tt)=.44(j 106 ft, the part of the diameter 
the third grinds oflf 

14. 61 n % of \it=\\ 7T==t'ie area after the fourth has 

ground off his his share. 

15. .-. V(ll 7r -T- 7r )=-IVf = L636634ft ' theradius/0 ' Then 

16. 2(^0 70)=2(5Vv-M)= 2(1.889822 ft. 

1.636634 ft)=.50e ! ,!76 ft,the part of the diameter the 
fourth grinds off. 

'. \7t f of 6^=11*^116 area after the fifth grinds 
off his share. 



18. 



. 
-*-n=* = 1.336306 ft, the radius HO. 



Then 



20. 



A , / _^ Ffl zv 7 )=2(1.636634ft- 

1.336306 ft)=.600f5/i ft., the part of the diameter the 

fifth grinds off. 
6^_e. f ^7r=|ft^=the area after the sixth grinds 

off his share. 

21. ../(Hir-r-Jr)==|VH'^ 49911 ft., the radius GO. Then 

22. 2(^0-GO)=2(iv/f-fV!)=2( 1.336306 ft 

.944911 ft.)=.78i790 ft, the part of the diameter the 
sixth grinds off . 

23. 2 X .944911 ft.=l. 889822 ft., the diameter of the part 

belonging to the seventh man. 

A M., having a woolen ball 2 feet in diameter, bored a 
hole 1 foot in diamer through the center. What is the volume 

Construction. Lz\.ABCDEF be a great circle of the ball 
-and let A CDF be a verticle section of the 
auger hole. Draw the diameter BOE 
.and the radius AC. Then the volume 
bored out consists of a cylinder, of which 
A CDF is a vertical section, and two 
spherical segments, of which ACB and 

~ are vertical sections. 

'1. BE=% feet=2/?, the radius of 

ther ball, and 
9. AO-\ foot=2r, the radius of the 

auger hole. FIG. 76 




FINKEL'S SOLUTION BOOK. 



II. 



. AF=2ij(R 2 

V=nr z 



3. 

4. . 

5. 

6. 2V / = 



7. 



=V3, the length of the cylinder. 

the volume of the cylinder, and 



. r(l |\/3) 3 =^7r(16- 9\/3), the vol- 
ume ot the two spherical segments. 
. F"+2F / =i^\/3-hV7r(16 9\/3)=:^7r(8 3>/3), 
=1.46809 cu. ft.=2536.85952cu. in. 

III. .-. The volume bored out is 2536.85952 cu. in. 

I. What is the diameter of the largest circular ring that cam 
be put in a cubical box whose edge is 1 foot? 

Construction. Let AB CD E be the cubical box. Let 7, K+ 
L, M, N, and P, be the middle points of the edge CF, GF, 
HA, AB, and BC respec- 
tively. Connect these points 
by the lines KI, KL, LM, 
MN, NP, and PI. Then 
IKLMNP is a regular 
hexagon, and the largest 
ring that can be put in the 
box will be the inscribed cri 
cle of the hexagon. 



edge of the cube. 




FIG. 77. 
Then 



TT . 

AN\/2=$\S2e, the 

side of the hexagon, 

4. MQ=ltML=ko 

5. OR= 

the radius of the circle. 

6. .-. 

14.6969382 in., the diameter. 

III. .. The diameter of the largest circular ring that can be 
put in a cubical box whose edge is 1 foot, is 14.6969382 in. 

I. A fly takes the shortest route from a lower to the opposite 
upper corner of a room 18 feet long, 16 feet wide, and 8 feet 
high. Find the distance the fly travels and locate the point 
where the fly leaves the floor. 



MENSURATION. 323 

Construction. Let FABE D be the room, of which AB is 
the length, AF the width, and 
AD the height; and let F be the 
position of the fly, and C the op- 
posite upper corner to which it 
is to travel. Conceive the side 
ABCD to revolve about AB 
until it comes to a level with the 
floor and takes the position of 
ABC'D'. Then the shortest 
path of the fly is the diagonal 
FC' of the rectangle FD'C'E, 
and P will be the point where FIG. 78. 

the fly leaves the floor. 

1. AB=a=.\% ft, the length of the room, 

2. AF=b=l ft., the width, and 
A.<]3. AD=A=& ft., the height. 

4. ^ZX=^+^ZX=J+>&=16ft.+8ft.=24ft. Then 
.5. FC'=\/(FD"*+D'C'*)= <t/[(-j-/O a + 8 ] 
IL < =V[(16+8) 2 -fl8 2 ]=30 feet, the length of the 

path of the fly. 

l. FD':D'C' -.'.AF-.AP, from the similar triangles 
C'D'F and PAF, or 




2. b+h :*::*: AP. Whence, AP= 

o-\-/i 

==12 feet, the distance from A to where the fly 
leaves the floor. 

J30 feet is the distance the fly travels, and [floor. 

' '1l2 feet is the distance from A to where it leaves the 

Remark. If we conceive the side BCHE to revolve about 
EH until it is level with the floor, the path of the fly will be 
FC" and the length of this is ^[( a +/i)*+t>*]. But \f[(a+k)* 
-|- 2 ] ^> \/[(-|-/z) 2 -|-tf 2 ], because, by expanding the terms under 
the radicals, it will be seen that the terms are the same, except 
'Zah and 2^, and since a is greater than , FC' is less than FC" '. 
When a=b, FC'=FC". 

I. How many acres are there in a square tract of land con- 
taining as many acres as there are boards in the fence inclosing 
it, if the boards are 11 feet long and the fence is 4 boards high? 

1. * -- =mumber of acres in the tract, the side being ex- 

pressed in rods. 

2. 4 Xl6X-^'<^ number of feet in the perimeter of the 

field. 



324 P^INKEL'S SOLUTION BOOK. 



3. V 4 X ri Xl6 ^ X5 ^ g "|=nu m ber of boards in the fence 

inclosing the tract. 

4. ... (. a ==4 l xlXJlW g = 24x^. Whence, 



5. 

6. .-. Vfe=3840 rods=12 miles. 
I?. .'. (3840) 2 -r-160=92160=number of acres in the tract. 

Ill, .'. There are 92160 A. in the tract. 

(Milne's Pract. Arith., p. 362, prob. 71.) 



SECOND SOLUTION. 

1. 16=number of acres comprised between two panels of 

fence on opposite sides of the field. 
2 lA.=43560sq. ft. 
3. 16 A.=16X43560 sq. ft.=696960 sq. ft. 

llft.=the width of this strip comprised between the two 

panels. 

... 12 mi. =63360 ft=696960-KLl, the length of the strip, 
which is the width of the field. 

6. 144 sq. mi.=( 12 ) 2 =the area of the field. 

7. 1 sq. mi.=640 A. 

8. 144 sq. mi.=144x640A=92160A. 

III. .-. There are 92160 A. in the tract. 

Explanation Since for every board in the fence there is an 
acre of land in the tract for 4 boards, or one panel of fence there 
would be 4 A. Now a panel on the opposite side of the field 
would also indicate 4 A. Hence, between two panels on oppo- 
site sides of the field there would be comprised a tract 11 ft. wide 
and containing 8 A. But this would make boards on the other 
two sides of the field have no value. Now the boards on the 
other two sides having as much value as the boards on the first 
two sides, it follows that we must take twice the area of the 
rectangle included between two opposite panels for the area com- 
prised between two opposite panels in the entire tract. Hence, 
between two opposite panels in the tract there are comprised 16 A. 
The length of this rectangle is 16 X43560-f-1 1=63360 ft =12 mi., 
which is the length of the side of the tract. 

In any problem of this kind, we may find the length of a side 
in miles, by multiplying the number of boards in the height of 
the fence by 33 and divide the product by the length of a board, 
expressed in feet. 




MENSURATION. 



325 



How many acres in a circular tract of land, containing as 
many acres as there are boards in the fence inclosing it, the fence 
being 5 boards high, the boards 8 feet long, and bending to the 
arc of a circle? 

Construction. Let C be the center of of the circular tract, AB 
=AC=1?, the radius, and the arc AjE>=8 feet. Then the area of 
.the sector is 5 A.=2 17800 sq. ft. 

'1. 5A.=5x43560sq. ft.=217800 sq. ft, the area of the 
sector ABC. 

2. ^(A^XAC)=^(SX^C)=4AC=ai-ea of the sector 

ABC. 

3. /. 44C=217800sq. ft. Whence, 

4. 4C=217800-5-4=54450 ft.=3300 rods, the radius of the 

circle. 

5. I-. 7t X (3300) 2 -f- 160=68062.5* = number of acres in 

tract. 

II. .-. There are 68062.5 TT A., in the tract. 

I. What is the length of a thread wrapped spirally around a 
cylinder 40 feet high and 2 feet in diameter, the thread passing 
around 10 times? 

1. 27rft.=ABCA (Pig. 19), the circumference of the cyl- 
inder 
II J 2. 4 ft.=40 ft.-T-10=^/ ?i , the distance between the spires. 

3. \/[(27r) 2 +4 2 ]=2\/[> 2 +4] .ft.=AJir, the length of 

one spire. 

4. .-. 10X2\/I> 2 +4]ft. =20V|> 8 +4] ft.=74.4838 ft, the 

^entire length of the thread. 

III. .-. The entire length of the thread=74.4838 ft 

Remark. Each spire is equivalent to the hypotenuse of a right 
angled triangle whose base is ABC A and altitude A P. This 
may be clearly shown by covering a cylinder with paper and 
tracing the position of the thread upon it Then cut the paper 
along the line APK o.r\d spread it upon a plane surface. AEP 
will then be seen to be the hypotenuse of a right-angled triangle 
whose base is A CBA and altitude AP. 

I. A thread passes spirally around a cylinder 10 feet high 
and 1 foot in diameter. How far will a mouse travel in unwind- 
ing the thread if the distance between the coils is 1 foot ? 

Construction. Let A CB /if be a portion of the cylinder and 
ADEFGKv. portioa of the thread. Let A be the position of 
the mouse when the unwinding begins, P its position at any 
time afterwards, A P N a portion of the path it describes, and PD 
the portion of the thread unwound. Diaw T DC parallel *o HB 
and draw QD and OC. Then 



FINKEL'S SOLUTION BOOK. 



II. 



10. 



?=2/?=l foot, the diameter of the cylinder. 
0=10 ft., the altitude. Let 
#=the angle AOC, 
s=AW, the length of a por- 
tion of the curve, 
x=0, and 
y=PL. Then 



GM=R cos 0, 

ML=fP=CP cos L CPI 
=Rd cos 7t l_ PCI) 



x=UM+ML= 7? cos 0+R0 
sin #, and 

IM= CM CI 
=R sinVCP cos 0= 



dx=R6 cos 6 dB, by differ- 

entiating in 10, 

dy=R8s\n6d6, by differ- 

entiating in 11. Now 




11. r= 

12. 
13. 

14. 
15. 

(R0 t iin0 dB)*}*=R cede 

=\RQ*. But 

16. 0=27T, when one spire is unwound, and 

17. #=10x27T=20/T, when the unwinding is complete. 

distance the mouse travels to unwind the thread. 
III. .-. The mouse will travel 989.96044 ft. to unwind the thread.. 

I. What is the length of a thread winding spirally round a 
cone, whose radius is R andaltitade , the thread passing round 
TZ times and intersecting the slant height at equal distances apart? 

Let Ploe any point of the thread, (x,y, z,) the co-ordinates of 
the point ;and, iet the angle PFC f (=DOC)=0, B0=a, the alti- 
tude, Z>0=, the radius of the base of the cone, and r=the 
radius of the cone at the point P. Then the equations of the. 



thread are : x-=~.r cos/9 ...... (1), yr sintf ..... (2), and *= 

____ (3). From the similar triangles DBF and DOB* 

f f) 

sYssv!?! 1 -- i ...C4). Now the distance between 
V 'InnJ 

P and its conoecutlvt position is 




f? 



MENSURATION. 327 



zr....(5). Substituting the value of r in (1) 
and (2), and differentiating, we 

have dx - - cos6-\- 
27rn\ 

(Inn 6) sin 'f\dO and dy= 






From (3), we have dz=-^^dO. 

Substituting these values of dx, 
dy, and dz in (5), we have 

Jo 2rc 



j? 2 [sing (2 n nB )cosV] 2 ) = 




where 7t=\/(a 2 -{-J? 2 ) , the slant height. 

NOTE. This solution was prepared for the School Visitor, by the author. 

I. A thread makes n equidistant spiral turns around a cone 
whose slant height is h, and radius of the base r. The cone 
stands on a horizontal plane and the string is unwound with the 
lower end in contact with the plane, the part unwound being 
always tense. Find the length of the trace of the end of the 
string on the plane. 

Let MH\>e the part unwound at any time, H being the point 
in contact with the cone, and BM=u, the trace on the plane up 
to this time. Put arc BE=x, AH=y, E being the point in the 
circumference of the base in the line AH. Let NI be the posi- 



328 FINKEL'S SOLUTION BOOK. 

tion of the string at the next instant, D and /being homologous 
points with E and H. Draw HK parallel to ED. Then h : 



DE : : AK: HK, or== (1). Now since the arc BE 



=x, is proportional to the distance the point 
of contact of the thread with the cone has 

x h y 
ascended, x'Ji y\ '.%7trn'.h, or- = ~. 

*/ ' ) sV/lt TT A 



.. >-=-==. tHHH (2). This is negative 

since y decreases as x increases. It is evi- 
dent from the figure that TT ^= = - 

IK dy h 

By similar triangles, IK'.HK \\ HE\ME, 

Jl/fT? f-TR' 

that is, form (1) and (2), we get- = ~F^ 

y dx ^y 




Therefore, ME= 

Put ME=t. 

. . . . (5). By similar figures r\ME\ \ED\ 



^_ 




From (3), put MP=, 

Equation (5) gives the entire addition to the line ME which 
-consists of NP+FD, since P*F=ME. Consequently, NP 
dt dx Znrn %nrn kitrn 

=lT y -dy=h^( h -^+- ir =^ry .... (7). Now MN* 

=MP*+NP* in the limit. Therefore ^lVy 



-^(fiy) 2 )? tne intregal of which is , the length of 



z 2 . Then u= f (7iz) 
a^hJo v 



the trace. Put // y=z 1 




a * -f- // - ) -i 

(11)- Write for 7z, its equal, 



MENSURATION. 329- 



4r 2^/^ 2 ^\ 

TT, in (11) and we have (12), u=- .-\-~i n n -- J 



This result is independent of h, the cone's slant height, but- 
involves ;z the number of turns of the thread. 

NOTE. This solution is by Prof. Henry Gunder and is taken from the 
School Visitor, Vol.9, p. 199. Prof. Gunder stands in the very front rank 
V Ohio mathematicians. He has contributed some very fine solutions to 
difficult problems proposed in the School Visitor and the Mathematical: 
Messenger. He is of a very retiring disposition and does not make any pre- 
tentions as a mathematician. But that he possesses superior ability along 
that line, his solutions to difficult problems will attest. Prof. Gunder was 
born at Arcanum, O , Sept. 15th, 1837. He passed his boyhood on a farm 
and it was while following a plow or chopping winter wood, that difficult 
problems were solved and hitherto unknown fields of thought explored. He 
became Principal of the Greenville High School in 1867. After seven years* 
work here, he became Superintendent of the Public, Schools of North Man- 
chester, Ind. After five years' work at this place he became Superintend- 
ent of schools of New Castle, Ind In 1890, Prof. Gunder was elected pro- 
fessor of Pedagogy in the Findlay, (Ohio) College. 

I. A woman printed 10 Ibs. of butter in the shape of a right 
cone whose base is 8 inches and altitude 10 inches. Having com- 
pany for dinner, she cut offa piece parallel to the altitude and con- 
taining \ of the diameter. What was the weight of the part cut off?- 

Construction. Let ABC G be the cone, A C the diameter 
and OG the altitude. Let E be the point 
where the cutting plane intersected the the di- 
ameter, F the corresponding point in the slant 
height, and DLFKB the section formed by the 
intersection of the cone and the cutting plane. 
Through F pass a plane parallel to the base 
AJBCand anywhere between this plane and 

the base, pass a plane NLMK. Then, 

! 

FIG- 82. 

1. A C=2^?=8in., the diameter of the base, 

2. OG=a=lQ in., the altitude, and 

3. OE==OCE C=R^A C=RIR=%R=\^ in.=c,the 

distance of the cutting plane from the altitude. Let 
4 GQ=x, the distance of the plane MLNK from the 
vertex G. Bv similar triangles, 

5. OC\OG\\EC\EF, or R\a'.\R^c\EF. Whence, 

6. EF= a ^ ~""^=6| in. By similar triangles, 

7. GO:OC::GQ:QM, or a\R\\x\QM. Whence, 

>= . Now, 
a 




330 FINKEL'S SOLUTION BOOK. 



II, 



9. area of LKM=arca of LQ KM area of LKQ. But 
10. area of LQKM=l(%LQ* cosr 1 ^^^^) 1 X 



cos "' and 

11. 



7?2 V 2 V^/rrX 

12. ,'. ^r^ of the segment LKM==~--~--co*-*l ^-} 



ment of volume of the part cut off. 



m 

14. .-. K= I (^r 

Jk 




= 19.6938154 10.0562976 + .6396202 = 



34.223792 cu. in., the volume of the part cut off. 

15. tf7r^=iXlOx42X*==53*r cu. in., the volume of 

the whole cone. 

16. 10 lbs.=the weight of the whole cone. Hence, by pro- 

portion, 

17. 537r cu. in. : 34.223792 cu. in. : : 10 Ibs. : ( ?=2.04258 Ibs. ) 

III. /. The weight of the part cut off is 2.04258 Ibs. 

I. After making a circular excavation 10 feet deep and 6 feet 
in diameter, it was found necessary to move the center 3 feet to 
one side; the new excavation being made in the form of a right 
cone having its base 6 feet in diameter and its apex in the surface 
of the ground' Reqired the total amount of earth removed. 



MENSURATION. 



331 



Construction, Let ABC F be the cylindrical excavation 
first made, A C the diame- 
ter, HO the altitude. Let 
A be the center of the con- 
ical excavation, GAH its 
diameter, and AP, an ele- 
ment of the cylinder, the 
altitude. Pass a plane at 
a distance x from O and 
parallel to the base of the 
excavation. Let figure II. 
represent the section thus 
formed, the letters in this 

section corresponding to FIG. 88. 

the homologous points in the base represented by the same let- 
ters in the base of the excavation. An element of the earth re- 
moved in the conical excavation is (area BAKGNB^dx. The 
whole volume removed in the conical part of the excavation is 




r 

Jo 



(area BAKGNB)dx. For let 



1. HO = a=lQ ft., the altitude of the excavation, 

2. HA=r=3> ft., the radius of the cylindrical and the 

conical parts. 



This is found from the proportion of 

-^-a-\-AI). Also 
~~(2rAI)AI=(rx-i-a Af)(rx+a+Af). Whence, 



3 A /? A 7\71, 
. ^~L j-j n. J. v . 

a 
similar triangles. 

4 /?/2 / rx-^-a 

^ ^f2-/2r 



r =rJ--=r( l~-^ }. Now 



10. area of BDAKGNB=1( area of BDAN-\-area of 

A^4G). But 

11. ^7r(r*x 2 -i-a 2 )= the area of the quadrant A^G, and 

12. area of BDAN=area of sector BAN-\-area of trian- 

gle HBAarea of sector BDAH. Now 

13. r0 of sector ^^A^^^^X^^sin-^^/-^-^^) 



*), and 



14. area of triangle ABH= 

XVC^ 8 * f )=(r i * 

15. area of sector BDAH= 



332 FINKEL'S SOLUTION BOOK 

16. .-. Area of BDAKGNB= C 1 \ \ 1 -^L 1t 4-' "* sin"* 





o-> 

S^2 V 2 ~2A 

r * ^""A 

r 2 cos- 1 ^ 1 ^1^ ( ^ = i7rr 2 + T^ 2 sin" 3 

V 22y i ' V 




=the volume of the conical part of the exca- 
vation. 

18. 7tar 2 =ihe volume of the cylindrical part. 

19. ,. 



r 2 =337.500554 cu. ft., the volume of the entire ex- 
cavation. 

III. /. The volume of the excavation = 



or 337.50055-f cu. ft, correct to the last decimal place. 

NOTE. This problem was proposed in the School Visitor by Wayland* 
Bowling, Rome Center, Mich. A solution of the problem, by Henry Gun- 
der, was published in Vol. 9, No. 6, p. 121. The solution there given is by 
polar coordinates. The editor'gives the answers obtained by the contribu- 
tors; viz., Mr. Dowling, H. A. Wood, R. A. Leisy, and William Hoover. 
Their answers differ from Mr. Gunder's and from each other. Mr. Gun- 
der's answer is 337.5-f-cu. ft., the same as above. There is a similar problem^ 
in TodhunteSs Integral Calculus, p. 190,prob. 29. 

I. A tree 74 feet high, standing perpendicularly, on a hill- 
side, was broken by the wind but not severed, and the top fell di- 
rectly down the hill, striking the ground 34 feet from the root of 
the tree, the horizontal distance from the root to the broken 
part being 18 feet, find the height of the stub. 

Construction. Let AD be the hill-side, AB the stump, 



MENSURATION. 333 

the broken part, and A C the horizontal line from the root of the 
tree to the broken part. Produce AB to E and draw DE paral- 
lel to A C. 

1. Let AB=x, the height of the stump. Then 

2. BD=1 ft. x=s #, the broken part, since AB-\-BD 

=74 feet. 

3. Let AD=a=M ft., the distance from the foot of the 

tree to where the top struck the ground, 

4. A C==18 ft., the horizontal distance from the foot 

of the tree to the broken part. 

5. x=AJ3, the height of the stump. Then 

6. BC=^(AB' 2 -\-A C 2 )=V(* 2 +^ 2 ) (1). In the sim- 

lar triangles BA C and BED, 

7. ^/(#2_|_2 ) . x . . s _ x . BE. Whence, 



II. 



11. ^^ 




= 2 . . . (5). Developing (5), we have 

14. 4 



15. 1161^ 4 91908* 3 -f 1959876* 2 25894080*+ 377913600 

=0 ..... (7), by substituting the values of a, b, 
and s in (6). 

16. .'. x =24 feet, the height of the stump, by solving (7) 

by Homer's method. 

III. .-. The height of the stump is 24 feet. 

NOTE. This problem was taken from the Mathematical Magazine^ Vol. 
I., No. 7, prob. 84. In Vol I., p. 184, of the Mathematical Magazine is a so- 
lution of it, given by C. H. Scharar and Prof. J. F. W. Sheffer The solu- 
tion there given is different from the one above. 

I. What is the longest strip of carpet one yard wide that can 
be laid diagonally in a room 30 feet long and 20 feet wide ? 

Construction. Let A BCD represent the room and EFGH 
the strip of carpet one yard wide placed diagonally in the room. 



334 FINKEL'S SOLUTION BOOK. 



II. 



1. Let A=a=3Q ft., the length of the room, 

2. C=t>=20 ft., the width, and 

3. HG=c=.% ft., the width of the carpet. Let 

4. BF=HD=x. Then 



6. 

7. AE=GC=ABEB=a 

similar triangles, 

8. EF:BF:: GF: GC, or 

9. c : x : : GF: a>J(c 2 x 2 . ) 

Whence, 



X 

Again, we have 

11. EF\BE\ : GF'.FC, or 

12. c:<J(c*-x*)::GF: b x. FIGi 85 . 




13. .-. LrJ?=- . . .(4). By equating G^in(3) and(4), 

- x ^ 



15. I >x _ x z ==a ^ c 2_ x 2)_ c ^ x 2 ...(6), by dividing (5) by 

cand clearing of fractions. 

16. c 2 bx 2x 2 =a*J(c 2 x 2 ) . ..(7), by transposing in(6). 

17. 4*4_4^ 3 +(0 2 + 2 4c 2 ) x 2 +2tc 2 x=c*(a*c* ) . . . - 

. . . (8), by squaring (7) and transposing and com- 
bining. 

18. 4* 4 80* 3 +l 264* 2 +360*= 8019 ____ (9), by restoring 

numbers in (8). 

19. .-. #=.2.5571+ft, by solving (9) by Horner's method. 

20. /V(^ 2 ^ 2 )=V(9^ 2 ) 1-5689 ft. Then, 

21. GC:=30 V(9 * 2 ) 28.4311 ft, and 

22. FC= 20 *=17.4429 ft. 

23. .-. G : ^^A/(^ 7 ^ 2 + G:C2 )= : V[(28.4311 

=33.3554 ft, the length of. the carpet. 

III. .'. The length of the strip of carpet is 33.3554 ft. 

I. What length of rope, fastened to a point in the circumfer- 
ence of a circular field whose area is one acre, will allow a horse 
to graze upon just one acre outside the field ? 

Construction. Let ABPC be the circular field and P the 
point in the circumference to which the horse is fastened. Let 
BP represent the length of the required rope. Draw the radius 
BO of the field and the line BC. Then 

r 1. 1 A.=160 sq. rd.=the area of the field ABPC, and 
2. BO=OP=R=\/lW+n)=\, the radius of 



MENSURATION. 335 

the circular field. Let 

3. 0=the angle ^/ > O=the angle OB P. Hence, 

4. TT 2#=the angle BOP. Now 

5. BP=APcosAPB=ZRcosO, the length of the - 

required rope. The 

6. area BP CD over which the horse grazes=area 

BE CDBareaBE CPB. 
But 

7. area of circle BECD= 



47T./? 2 cos 2 #, and the 

area BECP=^X(area of 
sector EPB-^-area of seg- 
ment BPH}. Now 




arc &JZ=X*tfcosy X FIG, 86. 

2.ffcos9x0=2./? 2 flcos 2 6>, and 

10. area of segment- BPH=area of sector BOP area 
triangle 
%[RXR(7t 20)] 1 

11. 



+ ^-2 6 sin2 61 - 

20cos2fl siH 2 2^]. 
12. . 



sin2#]. But 

13. 7T/? 2 =1A.=160 sq. rd.=the area of BPCDB, by the 

conditions of the problem. 

14. .-. 47r7? 2 cos 2 6> ^?2 

Whence, 



15. 4;r 7r2^cos2<9sin2^=:7r or 



16. 2?r--27rcos2# n 

17. . 

18. 2^ tan2#=27T, by dividing by cos2^. Whence, 

19. 0=51 16 r 24 X/ , by solving the last equation by the 

method of Double Position. 

20. .-. 

III. .'. The length of the rope is 8.92926+rods. 

I. If a 2-inch auger hole be bored diagonally through a 4-inch 
cube, what will be the volume bored out, the 'axis of the auger 
hole coinciding with the diagonal of the cube? 

Formula. V=r 2 ^( ite 2r^2), where is the edge. 



336 



FINKEL'S SOLUTION BOOK. 



Construction. Let AFGD be the cube and DF the diagonal,, 
which is also the axis of the auger hole. The volume bored out 
will consist of two equal tetrahedrons acd D and efg F plus 
the cylinder acdf, minus 6 cylindrical ungulas each equal to 
ace b. Pass a plane any where between e and , perpendicular 
to the axis of the cylinder, and let x be the distance the plane is- 
from D. Now let 



II. 



1. AHe=4: inches, the edge of the cube ; 

2. ZXF=\/3.y=4v'3, the diagonal of the cube; and 

3. r=\ inch, the ra- 

dius of the auger, 
or the radius of 
the circle acd. 

4. acad= dc = r\/3 
=V3, 

5. /te = iry6 = iV6, 

by the similar tri- 
angles dDc and 
HDc. 



9. 
10. 

11. 



13. 
14. 




5=iV2, the al- 
titude of the tetra- 
hedron acd D. 

.'. 2v = \(area of base X altitude) = 
-^r\/2-)=-|y'6r 3 =-^Y'6, the volume of the two tetrahe- 
drons, 

v / =7fr 2 X(^^2 times the altitude of acdD)= 
7rr 2 (e\/3--i>V 2 )= 7r (4A/3- -h/2), the volume of the 
cylinder acd /". 

fo=^r\/2, by similar triangles, not shown in the figure. 

^r\/2-f-^r\/2=r\/2=distance from D to where the 
auger begins to cut an entire circle. 

r %x\f1=versine of an arc of the ungulas at a distance 
x from D. 

12. 2rcos~~ 1 f y=an arc of the ungulas at a distance 



x from 
r 2 cos~ 1 



ment at a distance # from D. 



-i/*E*V 



MENSURATION. 



337 






15 V,the volume bored c,ut,=2^+^ / 6^ x/ = 

7T V /2 ) = r V3 ( ^ 



=16.866105 cu. in. 
III. .'. The volume bored out is 16.866105 cu. in. 
I. A horse is tethered to the outside of a circular corral. The 
length of the tether is equal to the circumference of the corral. 
Required the radius of the corral supposing the horse to have the 
libertj of grazing an acre of grass. 

Construction. Let AEFBK be the circular corral, AB the 
diameter, and A the 
point where the 
horse is tethered. 
Suppose the horse 
winds the tether 
around the entire 
corral; he will then 
be at A. If he un- 
winds the tether, 
keeping it stretched, 
he will describe an 
involute, APGH\ 
to the corral. From 
H' to H, he will de- 
scribe a semi-circle, 
radius AH'=AH= 
to the circumference 
of the corral. From 
H through G to A 9 
he will again de- 
scribe an involute. 

Then the area over which he grazes is the semi-circle HLH'-\- 
the two equal involute areas AFGHA and AKGH'A-\-\he area 
BFGKB. 

Let C be the center of the corral and also the origin of co-ordi- 
nates, A G the *-axis and P any point in the curve APGH'. 

1. Let #=the angle A CE that the radius CE perpendic- 
ular to PE, the radius of curvature of the curve 
APGH' , makes with the #-axis, 

<9 =the angle AFEBK that the radius CK makes 
with the #-axis when the radius of curvature PE has 
moved to the position KG\ 
RA C, the radius of the corral; 




FIG. 88. 



2. 



3. 



4. p=PE=arc AFE=RV, the radius of curvature of 



338 FINKEL'S SOLUTION BOOK. 

the involute ; 

5. *=C3/and 

6. y=PM, the co-ordinates of the point P\ and 

8. y =0, the co-ordinates of the point G. Then we have 

9. x=CM=lECD=PE(=arcAFJ5) cosZ/^P, 

=RS cos( i IEPi ECD}R cos(ar 0)=7?0 cos 
[^TT (n 6)] R cos(7r6)=ft6 cos -(%n- 8) 
Rcos(7t6)=R8cos8+R sin0 . . . . (1). 

10. y=PM=PI+IM(=DE}=PE sin^: PEI+ECx 

sin AECJD=RO sin(# ^7t)-\-R sin(?r ft)=R s'mtf 
RVsinV. . . . (2). When ^=^ =angle AFEBK 

11. * =CG=^costf +.## sin6> .... (3), and 

12. y =0=R sinV R6 cosV (4). Hence, from (4), 

13. =/?sin# -:-/?cos# =:tantf (5). Then, from (3) r 

IL ' ) - =Rcos0 




(6). Now 
15. BFGKB=V(\KGKKC sector 



16. 

=R*(%7iei).... (8), and 

17. HH'L^\n(AHy=\7t( f lnRY= : ln^R^ ....(9). Ad- 

ding (7), (8), and (9), 

18. /?2 6 i o _ /? 2( (9 ._ 7r)+ ^2( 87r i_^3) + 27r^2 =: 

7? 2 (7r+L 4 7T 3 ^^ t) 3 )=area over which the horse 
grazes. 

19. 1 A. =160 sq. rd*=43560 sq. ft.=the area over which. 

the horse grazes. 

20. .-. /? 2 (7T+y7r 3 ^^)=43560 sq. ft. Whence, 



22. B -=4.494039=264 37 X 18 x/ .35 by solving (5) by the 

method of Double Position. 
V 23 ;. 7?= 19 24738 ft, by substituting the value of 6 in(10). 

III. .-. The radius of the corral is 19-24738 ft. 

A 20-foot pole stands plumb against a perpendicular wall. A 
cat starts to climb the pole, but for each foot it ascends the pole 
slides one foot from the wall; so that when the top of the pole is 
reached, the pole is on the ground at right angles to the wall. 
Required the equation to the curve the cat described and the 
distance through which it traveled. 



MENSURATION. 



Construction. =Let A C be the wall, P the position of the cat 
at any time, and BC the position of the ladder at the same time. 
Draw AP and to the middle point D of AP draw BD. Then 



1. Let #C=20 ft=0, the length of the ladder, 

2. AP=r,the radius vector of the _____ 

curve the cat describes, and 

3. 0=the angle PAB. 

4. TT 20=the angle ABP, because 

the angle PAB=\he angle 
BPA. 



FIG. 89. 



II. 



III. 





6. AP=%r=AD=ABcos^ BAD 

= a cos2#cos#. 

7. .-. r= 2cos2#cos0, or 

8. r-}-2tf cos2#cos#=0, the equation of the curve de- 

scribed by the cat. 

1. Let 5=the distance through which the cat traveled. 

2. s=J\/(dr*+r*de*)=ZaJ* * ^(1 12 cos* 

44 cos 4 6 32 cos 6 0)d6, 



f\p 

. = a I 

Jo 

where 

/ 
. = / 

.70 



cos0 cos 2 0-|-4cos 3 0)</0. 



where 0=;r 20, 



cos0 2 



=1.1193 a, 

5. =22.386 ft, the distance through which the cat 
travels. 

r _|_2#cos2 #cos 0=0, is the equation or the curve, and 
22.386 ft.=the distance through which cat traveled. 



NOTE. The integration in this problem is performed by Cotes' Method 
of Approximation. 

I. Suppose W. A. Snyder builds a coke oven on a circular 
bottom 10 feet in diameter. While building it, he keeps one end 
of a pole 10 feet long, always against the place he is working 
and the other end in that point of the circumference of the bot- 
tom opposite him. Required the capacity of the oven. 

Construction. Let AB be the diameter of the base and CG 
the altitude. At a distance x from the base pass a plane inter- 
secting the oven in /^and E. Draw AE and A C. 



340 



II. 



FINKEL'S SOLUTION BOOK. 

f 1. AB= { 1R=\Q feet, the diameter of the base. 

2. A C=AJ5 < 2fi=lQ feet, by conditions of the problem 

3. CG=\/(AC 2 AG 2 )=(4:R 2 Ri) = R*]Z, the alti 
tude. 

4. 



=ZR*2fiX GH(=EI), because 
the ordinate of a semi-circle whose diameter is 2, 
From this, we find 

/?. Then 
A: 2 ) ^?] 2 , the 



area of the circle whose center is/. 

/rffys 
.-. l r = 7 

Jo 

/ 

Jo 



Q 




FIG 90. 



9. =^7T/?3(9V3 47T)=:^5 3 (9V3 47T), 

10. =i*rl25(9V3 4?r) =395.590202+ cu. ft. 
III. .-. The capacity is 395.590202 cu. ft. 

I. At each corner of a square field whose sides are 10 rods, a 
horse is tied with a rope 10 rods long; what is the area of th 
part common to the four horses? 

Construction. Let ABCD be the field and EFGH the 
common to the four horses. Join EF, 
FG, GH, and EH. Draw DKyzr 
peudicular to EF and draw DE and 
DF. Since AF=DE=>F= GB = 
CE, the triangles ADF and EDC 
are equilateral and, consequently, the 
angle A DE= / A D C Z ED C = 90 
60=30. Also the angle FDC= 
30 C . Hence, EDF=3>&. Now let 




II. 



1. AD=ED=a=\Q rods. 

2. Area of sector 



Then 



FIG. 91. 



arc 



3. Area of uiangle EDF=\EF^DK. But, 
4. 



XXII., = 
6. 



2^3), and 



of triangle 



, by formula of Prob. 



Hence, 

XW(2 V3)= i* : 



MENSURATION. 341 



7. /. Area of segment EF=^7ta' i ^a 2 =^a 2 (?r 3). The 

8. area of square EjFGH=EI?' i a 2 (2 y'3). Hence, 

9. area of the figure EFGH=a' i (^ y'3)+4x ^a 2 (7t 3) 

= 2 (^7r-[-l V3)=31.5147 sq. rd. the area common__ 
to the four horses. 

III. .-. The area of the part common to the four horses is 
31.5147 sq. rd. 

NOTE. This problem is similar to problem 348, School Visitor, to which 
a fine trigonometrical solution is given by Prof. E. B. Seitz. 

I. What is the length of the longest straight, inflexible stick 
of wood that can be thrust up a chimney, the arch being 4 feet 
high and 2 feet from the arch to the back of the chimney the 
back of the chimney being perpendicular? 

Construction. Let PDE C be a verticle section of the chim- 
ney, PB the height of the arch, PE the distance from the arch 
to the back of chimney, and APD the longest stick of wood that 
can be thrust up the chimney. 

1. Let PjB=a4: feet, the height of the arch, 

2. PE=b=2 feet, the width of the 

chimney. 

3. *=the length of the longest stick 

of wood, and 

4. fethe angle DA C. Then 

5. AP=PB cosec0= cosectf, 

7. .'. x=AP+PD=a cosectf +b 

sec 6 (1). Differentia- 

_ting(l), 

II., cos4!...(2)! n or 

9. a cos*0=Z>sin*0 ....(3), by FIG. 92. 

clearing of fractions and transposing in (2). 

10. .'. ?^!==tan30=?. Whence, 




11. tan #=;*]-. From (3), we may also have 

12. cot# =!H Now, from trigonometry, 

13. v /(l_|-tan 2 )=sec0, and 

14. v / (l-f cot2/9 )= cosec ^- Hence, by substituting in (1), 

15. xai/ 



342 FJNKEL'S SOLUTION BOOK. 

III. .- The length of the longest stick is 8.323876+ft 

I. A small garden, situated in a level plane is surrounded by 
a wall having twelve equal sides, in the center of which are 
twelve gates. Through these and from the center of the garden 
12 paths lead ofFthrough the plane in a straight direction. From 
a point in the path leading north and at a distance of 4 furlongs 
47_i^ yards from the center of the garden, A. and B. start to travel 
in opposite directions and at the same rate. A. continues in the 
direction he first takes; B., after arriving at the first road (lying 
east of him) by a straight line and at right angles with it, turns so 
as to arrive at the next path by a straight line and at right angles 
with it and so on in like manner until he arrives at the same 
road from which he started, having made a complete revolution 
around the center of the garden. At the moment that B. has 
performed the revolution, how far will A. and B. be apart? 

Let O be the center of the garden, A the point in the path 
leading north from which A. and B. start, C, D, E, F, G, H, f, 
K,L, M, N, P, the points at which 
B. strikes the paths. The triangles 
OCA, ODC, OED, OPE, &c., are 
right triangles, OCA, ODC, OED, 
OEF, &c., being the right angles. Let 
S in the prolongation of A C denote the 
position of A., when B., arrives at P. 
It is required to find the distance AS. 
Let OA=a=4 furlongs, 47^ \ y d -> Ps 
=A C+ CD+DE+ . . . +NP=x, AS 
=y, AP=z, #=12, the number of paths 
and LAOC= LCOD= LDOE= ____ 
/AW ) =360-:-72=30 . Then from the _ 
right triangles we have OC=OAX FIG, 93. 




OD=OC cosCO7}=a cos 2 
cos DOE=a cos 3 #, OP=ONcos NOP=a cos^; A C= 
sinAOC=asin6, C>=OCsinCOZ)=a sin0 



.-. z=OA OP--=a(\ 
tfsin(9 cos 2 #-(- ..... -\- 
cos 3 #-f- ..... -(-cos^ 1 
acot%0(lcos n 0). Hence, since PAS=(W+e ), we have 

X 



yd., nearly. 



NOTE This problem was proposed in the School Visitor, by Dr. N. R. 
Oliver, Brampton, Ontario. The above elegant solution was given by Prof. 
E B. Seitz, and was published in the School Visitor, Vol 3, p. 36. 



MENSURATION. 34S 

I. A fox is 80 rods north of a hound and runs directly east 
350 rods before being overtaken. How far will the hound run 
before catching the fox if he runs towards the fox all the time, 
and upon a level plain? 

Construction. Let C and A be the position of the hound and 
fox at the start, P and m corresponding positions of the hound 
and fox any time during the chase, and P' and n their positions 
the next instant, B the point where the hound catches the fox 
and CPP' B the curve described by the hound. Join m and P, 
and n and P'jthey are tangents to the curve at P and P' . Draw 
Pd and P'e perpendicular to AB , mo perpendicular to /",. 
and P'p perpendicular to Pd. 



2. AB=b= 350 rds., 
4! JBd^ 




6. arc CP=s, 

7. curve CPB=s^ and 

8. r=ratio of the FIG 94. 

hound's rate to the fox's. Then we have 

10! ed=P'p=dy, 

11. PP'=ds, 

12. noPP'=diu (1), and 

13. srx .... (2). From (2), we have, by differentiation,. 

14. ds=rdx. Whence, 

15. -j-=~. From the similar right triangles PpP / and mon, 

we have 

16. PP / '. ?nn\\pP / \ mo^ or ds ; dx '.'. dy : mo. Whence, 

_^ _dy ^_^^dx_l 

11.1 

dy 

18. -^-dsdw, or 

r 

19. ^y r 2^ x==rc ^ w ^j Integrating (3), 

20. y r 2 x=rw-\-C. . . .(4). But, since when #=0, 

and w=a, 

21. 0=rfl+C. Whence, C= 

22. /. y r z x-=rw-\- C=rw ra .... (5). When x=b, j 

=^,and w=0, and (5) becomes 

23. b ^6= ar, or r*b ra=b. Whence, 

94 r i ?, 1 

Zi^t. / ,' 1 5 



17. mo=- ; = , since- T -=-. Substituting: in (1). 
ds r ds r 




344 



FINKEL'S SOLUTION BOOK. 



26 - 

27. 



28. r=|(0-|-tf 2 +4 2 ). But 

29. rb=s l ^ what (1) becomes when =#. 

.30. .*. 5 1 =(+Vtf 2 -r-4 2 y== ; ^2.2783 rods, the distance the 
hound runs to catch the fox. 



NOTE. This solution is substantially the same as the one given by the Late 
Professor E. B. Seitz, and published in the School Visitor, Vol.2 V< p. 201. 
The path of the hound is known as the "Curve of Pursuit." 

I. A ship starts on the equator and travels due north-east at 
^all times ; how far has it traveled when its longitude, for the 
first time, is the same as that of the point of departure? 

Let B be the point of the ship's departure, B 'PJVits course, P 
its position at any time and TV^its position at the next instant. 
Then PN is an element of the curve of the ship, which is known 
as the Loxodrome, or Rhumb line. Let 6=jBI?=\\\e longitude 
of the point P, 0=PJ 7 =the corresponding latitude, (#,jy, z) 
the rectangular co-ordinates of P , and ^.=^7r=the constant 
angle PNQ. 

Then we have for the 
equations of the curve, 
x=PGcos6 
=rcos0cos# .... (1), 
y=P G sin 0=r cos0 X 
sin# .... (2), and 
z=r sin0. . . .(3), where 
r is the radius of the 
earth. Now an element 
of a curve of double 
curvature, referred to 
rectangular co-ordinates 

is V(<fcM-< 

.-. PN=ds 




...(4). Differentiating 

<i), (a), d (*),.. 

cos0sin#df#), f/G- 55. 

dy= r(sin#sin0</0 cos0cos&/0), and dz=rcos (f>d(f). 

stituting these values in (4), ds=r<\/[cos' 2 (/)d0' 2 -\-(s'm 



Sub- 



.... (4). Now 



and N= 



Substituting the value of cos0^i9 i 



MENSURATION. 345- 



=ta.n(p, or 




==tan9log a [tan(i?r+i0)] or ^ cot( P=tan(i7r+|0) . . . (8). 
Whence, 0=2tan~ 1 (e dcoi( P) \it . When 6^=2^ and cp=l ?r, 




the distance the ship travels. 

The rectangular equations of the Loxodrome are V(^ 2 
j atan- i !_j_ 6r atair 1 ! ( = 2r, and x 2 +y*+z 2 =r 2 , where = 

The last equations are easily obtained from the figure. The 



first is obtained as follows: From (1) and (2), we find 



v 
~ J - 



X 



also, # 2 -f-J 2 =^ 2 cos 2 or cos0=-V(^ 2 +jK 2 ). From (8), we get 

J^0 COS< ^ Whence, <0 *<P+e<> *<P * 

l+sin0 1+V( Icos 2 0) 

os 2 0)=cos0. Transposing eOcoW, squaring, and reducing, 
we have cos0(^ col< ? > -f-^ cot( ? > )=2. Substituting the value cos0, 

( y y ) 

and #, we have \/( x 2 -\-y^) \ e ai ~ 1 K-\-e~ ai&u ~ 1 v > =2r. 

NOTE. This solution was prepared by the author for problem 1501> 
School Visitor, but it was not published because of its difficult composi- 
tion. 



346 FINKEL'S SOLUTION BOOK. 



PARALLELOGRAMS. 

1. Find the area of the parallelogram ABCD; given AC=7 ft. 2 in., and 
the perpendicular from B on AC 3 feet. [See Fig. 4, p. 198.] 

2. Find the area of a parallelogram in which one side is 4 ft. 3 in., and 
the perpendicular distance between this and the opposite side is 4 feet. 

3. The area of a parallelogram is Yiy 2 acres, and each of two parallel 
sides is 42 chains ; find the perpendicular distance between them. 

4. Find the area of a rhombus, a side of which is 10 feet and a diagonal 
of 12 feet. [The diagonals of a rhombus bisect each other at right angles.] 

5. Find the area of a rhombus whose diagonals measure 18 feet and 24 
feet. 

6. A field in the form of a rhombus, whose diagonals are 2870 links and 
1850 links; find the rent of the field at $5 per acre. 

7. The diagonals of a parallelogram are 34 feet and 24 feet, and one side 
is 25 feet ; find its area. 

8. Find the cost of carpeting a room, 30 feet long and 21 feet wide, with 
carpet 2 feet wide at 80 cents per yard. 

9. How many square yards are there in a path, 4 feet wide, surrounding 
a lawn 24 yards long and 22 yards wide ? 

10. How many yards of paper, 20 inches wide, will be required to paper 
the walls of a room, 16 feet by 14 feet by 9 feet, allowing 8 inches for a base- 
board at the floor and 12 inches for border at the ceiling? 

11. The perimeter of a rectangle is 56 feet; find its area, if its length is 
3 times its breadth. 

12. What is the area in acres of a square whose perimeter is such that 
it takes 12 minutes to run around the square, at the rate of 5)4 miles per 
hour? 

13. Cut a rectangular board, 16 feet long and 9 feet wide, into two pieces 
in such a way that they will form a square. 

14. How many feet of framing, 4 inches wide, will it take to frame a 
picture, 3 feet by 2 feet ? Ans. 6 ft. 4 in. 

15. A sheet of galvanized iron, 50 inches wide, is placed against the top 
of a wall, 6 feet high, while the lower edge is 5 feet 5 inches from the foot 
of the wall ; find the area of the sheet of iron. Ans. 4850 sq. in. 

16. Allowing 8 shingles to the square foot, how many shingles will it 
take to roof a barn which is 40 feet long and 15 feet from the comb to the 
eaves ? Ans. 9600 shingles. 

17. The area of a square is 169 sq. ft. ; find its perimeter, in chains. 

18. What is the side of a square, of which the number expressing its 
area in square feet is equal to the number expressing its perimeter in 
yards? Ans. 1^ feet. 

19. What is the area of a path a yard wide, running diagonally across a 
square lawn whose side is 30 feet? Ans. 648[20 l /~2 1] sq. in. 

20. What is the area of a square whose diagonal is 12 feet ? 

Ans. 72 sq. ft. 

21. What is the area of a square whose diagonal is 5 feet longer than 
its side? Ans. 25 (3+2 /~3) sq. ft. 

TRIANGLES. 

1. A man travels 20 miles north, then 15 miles due east, finally 28 miles 
due south ; what is the distance from his starting point ? Ans. 17 mi. 

2. A ladder, 50 feet long, is placed so as to reach a window 48 feet high, 



PROBLEMS. 



347 



and on turning the ladder over to the other side of the street, it reaches a 
point 14 feet high. Find the breadth of the street. 

3. The hypotenuse of a right-angled triangle is 55 feet and the base is 
^ of the altitude. Find the two sides. 

4. The hypotenuse of a right-angled triangle is 13 feet and the sum of 
the sides containing the right angle is 17 feet. Find these sides. 

5. In a right-angled triangle the area is half an acre, and one of the 
sides containing the right angle is 44 yards ; find the other side in yards. 

6. Find the area of a triangle whose sides are 21 feet, 20 feet, and 13 
feet, respectively. Also 21 feet, 17 feet, and 10 feet. 

7. In a right-angled triangle the sides containing the right angle are 30 
feet and 40 feet. Find the length of a perpendicular drawn from the right 
angle to the hypotenuse. 

8. The perimeter of a triangle is 48 feet. If one side is 10 feet and the 
area is 84 square feet, find the two remaining sides. 

9. The area of an equilateral triangle is 30 square feet. Find the length 
of a side. What is the side of a square of equal area ? 

10. The sides of a triangle are proportional to 3, 4, and 5. If the perim- 
eter is 84 feet, find the sides and the area. 



TRAPEZOIDS. 

1. The parallel sides of a trapezoid are 18 feet and 24 feet, and the alti- 
tude is 8 feet ; find the area. 

2. The parallel sides of an isosceles trapezoid are 16 feet and 20 feet, 
and the non-parallel sides are 10 feet each ; find the area of the trapezoid. 

3. The line joining the middle points of the non-parallel sides of a 
trapezoid is 12 feet, and the altitude is 8 feet ; find the area of the trape- 
zoid. Ans, 96 sq. ft. 

TRAPEZIUMS AND IRREGULAR POLYGONS. 



1. In the trapezium ABCD, AB=3& in., BC= 17 in., CZ>=25 in., 

28 in., and the diagonal D=26 in.; find its area. Ans. 540 sq. in. 

2. In the quadrilateral ABCD, the diagonal AC=18 in., and the per- 
pendicular on it from B and D are 11 inches and 9 inches respectively; 
find the area of the trapezoid. Ans. 180 sq. in. 

3. In the trapezium ABCD, the diagonals AC and BD are perpendicu- 
lar to each other and measure 16 feet and 2^ feet respectively ; find the 
area. Ans. 2 sq. yds. 

4. Find the area of the trapezium ABCD, in which the angles ACand 
CD A are right angles, and AB is 15 feet, BC is 20 feet, and CD is 7 feet. 

Ans. 234 sq. ft. 

5. Find the area of the quadrilateral ABCD, having given that the 
angle ABC is 60, ADC is a right angle, ^=13 chains, BC=IZ chains, 
and CD=12 chains. Ans. 10.31 acres. 

6. The area of a trapezium is 4 acres and the two diagonals measure 
16 chains and 10 chains respectively; at what angle are the two diagonals 
inclined to each other ? Ans. 30. 

Hint. Let / be the intersection of the diagonals. Then 1 from B on AC=BI sin. 
LBIC,\. from D on AC=DI sin.^>JA=(/_CIB.) 

. . area of ABCD=y 2 AC[BI sin. BIC+DI sin. CIB]=y 2 ACy.BDX$\n. LBIC. 

7. Find the area of the polygon ABCDEF, if AD=1G75 links, 1 FP 
from .Ton AD=8bQ links, 1 BQ from B on ^Z?=-200 links, 1 CS from Con 
on AD=50Q links, 1 ER from E on AD=25Q links, AP=9QO links, AQ= 
1040 links, AR=12W links, and ^5=1380 links. Ans. 9.03625 acres. 



348 'FINKEL'S SOLUTION BOOK. 

8. Find the area of the field ABCDEF, if ^C=2900 links, C 
links, ^^=3600 liiiks, 1 BX from B on AC=4QQ links, i >Kfrom D on 
C=4QQ links, and 1 FZ from F on ^"=950 links. Ans. 63 A. 3 r. 24 p. 

9. Find the area of the polygon ABCDE, if AB=IZ inches, L^BC a 
right angle, BC= in., CZ?=14 in , AD=\5 in., /_ADE a right angle, and 
DJ5=8 in. ^f^. 1 sq. ft. 30 sq. in. 

REGULAR POLYGONS. 

1. Within a given regular hexagon, drawn on a side of 10 inches, a 
second hexagon is inscribed by joining the middle points of the sides 
taken in order. Find the area of the inscribed figure. Am. 194.85 sq. in. 

2. Find the area of a regular pentagon on a side of 10 inches. 

Ans. 172.04 sq. in. 

3. Find the area of a regular decagon on a side of 4 inches. 

Ans. 123.1 sq. in. 

4. Find the area of a regular heptagon inscribed in a circle, radius 6 

n 360 

inches. [Area of a regular #-side in terms of the radius is-- sin. 



5. Find the area of a regular heptagon circumscribing a circle whose 
radius is 12 inches. [Area of a regular -side circumscribing a circle in 

180 
terms of the radius is n tan.-^ (r) 2 ]. 

6. A regular octagon is formed by cutting off the corners of a square 
whose side^_12 inches. Find the side of the octagon. [Side of octagoa 
= i^a(2 v/ 2), where a is the side of the square]. 

7. The area of a dodecagon is 300 square inches ; find the radius of the 
circle circumscribed about it. 

8. Find the area of the circular ring formed by the inscribed and cir- 
cumscribed circles of a regular hexagon whose side is 20 inches. Show 
that for a given length of side, the area of the ring is the same whatever 
the number of sides of the regular polygons. 

9. What is the area of a path 3 feet wide, around a hexagonal enclosure 
whose side is 14 feet ? Ans. 283 . 17 sq. ft 

10. Find the area of the square formed by joining the middle points of 
the alternate sides of a regular octagon, whose side is 8 inches. 

Ans. 186.51 sq. in. 

11. The difference between the area of a regular octagon and a square 
inscribed in the same circle is 82.8 square inches. Bind the radius of the 
circle. [Take j/~2=1.414]. Ans. 10 inches. 

12. In a circle of a radius 10 inches a regular hexagon is described ; 
in this hexagon a circle is inscribed ; in this circle a regular hexagon is 
inscribed : and so ad infinitum. Find the sum of the areas of all the hex- 
agons thus formed. Ans. 1039.23 sq. in. 

13. In the last example, let the radius be r and the number of sides of 
the polygon n ; find the sum of the areas of all the circles formed. 

180 

Ans. K r 2 cosec. 2 -- 
n 

14. In a triangle whose base is 15 inches and altitude 10 inches a square 
is inscribed. Find its area. 

CIRCLES. 

1. The driving-wheel of a locomotive engine 6 feet 3 inches in diam- 
eter, makes 110 revolutions a minute ; find the rate at which it is traveling. 

Ans. 24.54 miles per hour. 



PROBLEMS. 



349 



2. If the driving-wheel of a bicycle makes 560 revolutions in traveling 
a mile, what is its radius? [Take *=%}]. Ans. \V Z feet. 

3. Find the area of a walk 7 feet wide, surrounding a circular pond 252 
in diameter. [Take *=3$]. Ans - 539 s q- ft - 

4. A wire equal to the radius of a circle is bent so as to fit the circum- 
ference. How many degrees in the angle formed by joining its ends with 
the center of the circle ? [Take * =3 . 14159265]. Ans. 57 . 2957795. 

Definition. The angle subtended at the center of a circle by an arc 
equal in length to the radius is called a radian. 

5. A wire is bent into the form of a circle whose radius is 30 inches. 
If the same wire be bent into the form of a square, what would be the 
length of its side ? 

6. A circle and a square have the same perimeter. What is the differ- 
ence between their areas ? 

7. Two tangents drawn from an external point to a circle are 21 inches 
long and make angles with each other of 90. Find the area of the circle. 

8. A bicycle driving-wheel is 28 inches in diameter, the sprocket-wheel 
has 17 sprockets, and the rear sprocket-wheel 7 sprockets ; what is the gear 
of the wheel ? 

Hint. One revolution of the sprocket-wheel makes 17-r-7=J^!!- revolutions of the rear 
sprocket-wheel, or JJZ. revolutions of the driving wheel, the rear sprocket-wheel and the 
driving-wheel being rigidly connected. . . 7TX28X-L7-=7rX68 inches, the distance trav- 
eled in one revolution of the sprocket-wheel. 68 inches is the gear of the wheel. Gear 
= ( n-z-m )>, where is the number of sprockets in the sprocket-wheel, m the number of 
sprockets in the rear sprocket-wheel, and D the diameter of the driving-wheel in inches. 

9. (a) What is the gear of a bicycle whose driving-wheel is 30 inches in 
diameter, whose sprocket-wheel has 19 sprockets, and whose rear sprocket- 
wheel has 6 sprockets? (b) How many revolutions of the sprocket-wheel 
will be required to travel a mile ? Ans. (a) 95 inches. 

10. What is the distance from the center of a chord 70 inches long in a 
circle whose radius is 37 inches ? Ans. 12 inches. 

11. In a circle whose radius is 9 inches, the chord of half an arc is 12 
inches ; find the chord of the whole arc. Ans. 17.89 inches. 

12. The length of an arc of a circle is 143 inches and its central angle is 
9 6'; find the radius of the circle. Ans. 900 inches. 

13. In a circle of a radius of 37 inches, find the length of the minor arc 
whose chord is 24 inches. Ans. 24.44 inches. 

14. The radius of a circle is 21 inches ; find the length of an arc which 
subtends an angle of 60 at the center. 

15. The radius of a circle is 9 feet 4 inches ; what angle is subtended 
at the center by an arc of 28 inches ? 

16. The chord of an arc is 48 inches and its height is 7 inches ; find the 
length of the arc. [Arc=^(8<5 a), where b is the chord of half the arc 
and a is the chord of the whole arc.] Ans. 50% inches. 

17. In a circle whose diameter is 72 inches, find the length of the arc 
whose height is 8 inches. 

18. Find the area of a sector of a circle whose radius is 21 inches and 
the angle between the radii 40. 

19. Find the area of the sector of a circle having given the arc 32 inches 
and the radius 17 inches. Ans. 272 sq. in. 

20. Angle of a sector is 36 and its area is 385 square feet; find the 
length of its arc. Ans. 22 feet. 

21. Find the area of a segment cut off by a chord whose length is 14 
inches from a circle of a radius of 25 inches. Ans. 9.37 inches. 



350 FINKEL'S SOLUTION BOOK. 

32.* A regular pentagon is inscribed in a circle of a radius 10 inches ; 
find the area of a minor segment cut off from the circle by one of its sides. 

Ans. 15.27 sq. in. 

33. Find the area of a segment whose chord is 30 inches and height is 
8 inches. Ans. 168.16 sq. in. 

34. Find the area of a circle inscribed in a sector whose angle is 120 
and whose radius is 10 inches. 

35. A line AB is 20 inches long, and C is its middle point. On AB, AC, 
and CB semicircles are described. Find the area of the circle inscribed in 
the space inclosed by the three semicircles. Ans. r=3^ inches. 

36. Two equal circles, each of a radius 9 inches, touch each other exter- 
nally, and a common tangent (direct) is drawn to them ; find the area of the 
space inclosed between the circles and the tangent. Ans. 7 . 53 sq. inches. 

37. Three circles of radius 3 feet are placed so that they touch each 
other ; find the area of the curvilinear space inclosed by them. 

Ans. 207 sq. in. 

38. From the angular points of a regular hexagon, whose side is 10 
inches, six equal circles, radii 5 inches, are drawn ; find the area of the 
figure inclosed between the circles. Ans. 50(3V 3^) sq. in. 

39. Two equal circles of radius 5 inches are described so that the center 
of each is on the circumference of the other ; find the area of the curvi- 
linear figure intercepted between the two circumferences. 

Ans. 30.71 sq. in. 

40. Two equal circles of radius 5 inches intersect so that their com- 
mon chord is equal to their radius ; find the area of the curvilinear figure 
intercepted between the two circumferences. Ans. 4.53 sq. in. 

41. Three circles, radii 10, 12, and 16 inches respectively, touch each 
other; find the radius of a circle touching the three circles. [See Prob. 
CLXVL] 

SIMILAR AREAS. 

[See principle on p. 309.] 

1. The sides of a triangle are 21, 20, and 13 inches; find the area of a 
similar triangle whose sides are to the corresponding sides of the first 
as 25: 3. 

2. In a survey map an estate of 144 acres is represented by a quadrilat- 
eral, ABCD. The diagonal, AC, is 6 inches, and the perpendiculars from 
B and D on AC are 1.8 inches and .9 inches respectively. On what scale 
was the map drawn ? Ans. 6 inches to the mile. 

3. A man 6 feet in height, standing 15 feet from a lamp-post, observes 
that his shadow cast by the light at the top of the post is 8 feet in length ; 
how long would his shadow be if he were to approach 8 feet nearer to the 
post ? Ans. 2 ft. 4 in. 

4. A man, wishing to ascertain the width of an impassable canal, takes 
two rods, 3 feet and 5 feet in length. The shorter he fixes vertically on 
one bank and then retires at right angles to the canal, until on resting the 
other rod vertically on the ground he sees the ends of the two rods in a 
line with the remote bank ; if the distance between the rods is 60 feet, 
what is the width of the canal ? Ans. 90 feet. 

5. A man wishing to find the height of a tower, fixes a rod 11 feet in 
length vertically on the ground at a distance of 80 feet from the tower. 
On retiring 10 feet further from the tower he sees the top of the rod in 
line with the top of the tower. If the observer's eye is 5^ feet above the 
ground, find the height of the tower. Ans. 55 feet. 

6. A triangle ABC is divided into two equal parts by a straight line 
XY t drawn parallel to the base BC. If A=WO inches, find AX. 



PROBLEMS. 351 

7. In a given triangle a triangle is inscribed by joining the middle 
points of the sides. In this inscribed triangle another similar triangle is 
inscribed, and so on. What fraction of the given triangle is the area of 
the sixth triangle so drawn ? 

8. (a) In a given square whose side is 16 inches a square is inscribed by 
joining the middle points of the sides of the given square ; in this inscribed 
square a square is inscribed in like manner, and so on ; find the area of the 
fifth square, (b) If the process be continued ad infinitum what is the sum 
of the areas of all the squares ? 

9. In a circle of a radius of 32 inches an equilateral triangle is in- 
scribed, and in this triangle a circle. In this circle an equilateral triangle 
is again inscribed, and in the triangle a circle, and so on. If the process is 
continued, find the area of the fourth circle and find which of the circles 
has an area of 3 sq. in.? Ans. 50f sq. in.; the sixth. 

10. A field of 9 acres is represented in a plan by a triangle whose sides 
are 25, 17, and 12 inches. On what scale is the plan drawn and what length 
will be represented by 80 inches? Ans. ,-^; 1 mile. 

12. The following is used by lumbermen in finding the diameter of 
trees at any height above the ground : If the tree casts a definite shadow 
on a horizontal plane, stand on the edge of the shadow and observe where 
the line of light from the sun to your eye strikes the tree. Then measure 
the shadow of the tree at the point where the shadow of your head strikes 
the ground. The width of the shadow is the diameter of the tree at the 
point where the line of light from the eye to the sun strikes it. What 
principle is involved ? 



1. Find the surface of a rectangular solid whose length is 12 feet, 
breadth 5 feet 4 inches, height 5 feet 3 inches. 

2. Find the cost of papering the four walls of a room whose length is 
20 feet 6 inches, breadth 15 feet 6 inches, and height 11 feet 3 inches, at 3d. 
a square yard. 

3. A rectangular tank is 16 feet long, 8 feet wide, and 7 feet deep ; how 
many tons of water will it hold, a cubic foot of water weighing 1,000 oz.? 

4. The surface of a rectangle is 1,000 sq. in.; if its length and breadth 
are respectively 1 ft. 3 in. and 1 ft. 2 in., find its height. 

5. The dimensions of a rectangular solid are proportional to 3, 4, and 5. 
If the whole surface contains 2,350 sq. in., find the length, breadth, and 
height. 

/#/. 2,350-5-[2(3X4) +2(3X5) +2(4X5)]=25, the greatest common divisor of the three 
dimensions. 

6. The whole surface of a rectangular solid contains 1,224 square feet, 
and the four vertical faces together contain 744 square feet. If the height 
is 12 feet, find the length and breadth. 

7. Find the surface and volume of a cube whose diagonal is 2 feet 6 
inches. Ans. 12^ sq. ft. ; 3 cu. ft. 12 cu. in., nearly. 

8. Find the edge of a cubical block of lead weighing one ton, having 
iven that a cubic foot of lead weighs 709^ Ibs. Ans. 17 .60+ inches. 

9. The edges of a rectangular block of granite are proportional to 2, 3, 
id 5, and its volume is 101 cu. ft. 432 cu. in. ; find its dimensions. 

Ans. 3 ft. ; 4 ft. 6 in. ; 7 ft. 6 in. 

10. The diagonal of a rectangular solid is 29 inches, and its volume is 
4,032 cu. in. ; if the thickness is one foot, find the length and breadth. 

Ans. 21 in. and 16 in. 



given 

9. 
and 5 



352 FINKEL'S SOLUTION BOOK. 



PRISMS. 

1. A right prism stands upon a triangular base, whose sides are 13, 14, 
and 15 inches. If the height is 10 inches, find its volume and whole sur- 
face. Ans. 840 cu. in. ; 4 sq. ft. 12 sq. in. 

2. The weight of a brass prism standing on a triangular base is 875 Ibs. 
If the sides of the base are 25 in., 24 in., and 7 in., find the height of the 
prism, supposing that 1 cu. ft. of brass weighs 8,000 oz. Ans. 3 ft. 

3. Water flows at the rate of 30 yards per minute through a wooden 
pipe whose cross-section is a square on a side of 4 inches. How long will 
it take to fill a cubical cistern whose internal edge is 6 feet? 

Ans. 21f min. 

4. Find the volume of a truncated prism ( that is the part of a prism 
included between the base and a section made by a plane inclined to the 
base and cutting all the lateral edges), whose base is a right triangle, base 
3 feet, and altitude 4 feet, and the three lateral edges 3 feet, 4 feet, and 5 
feet respectively. [Formula. J /r =^A(e 1L -{-e s -\-e 3 ) J where A is the area of 
the base and e lt e s , and e s the lateral edges.] 

CYLINDERS. 

1. How many cubic yards of earth must be removed in constructing a 
tunnel 100 yards long, whose section is a semi-circle with a radius of 
10 feet? 

2. Find the convex surface of a cylinder whose height is three times its 
diameter, and whose volume is 539 cubic ihches. 

3. The cylinder of a common pump is 6 inches in diameter ; what must 
be the beat of the piston if 8 beats are needed to raise 10 gallons? 

Ans. 12 in. 



4. A copper wire ^ inches in diameter is evenly wound about a cylin- 
der whose length is 6 inches and diameter 9.9 inches, so as to cover the 
convex surface. Find the length and weight of the wire, if 1 cu. in. of 
copper weighs 5.1 oz. Ans. 1,885 in., nearly; 75.5 oz. 

5. A cubic inch of gold is drawn into a wire 1,000 yards long. Find the 
diameter of the wire. Ans. .006 in. 

6. The whole surface of a cylindrical tube is 264 square inches ; if its 
length is 5 inches, and its external radius is 4 inches, find its thickness. 
[Use7r=3f.] Ans. I in. 

7. If the diameter of a well is 7 feet, and the water is 10 feet deep, how 
many gallons of water are there, reckoning 7^ gallons to the cubic foot? 

PYRAMIDS AND CONKS. 

1. Find the entire surface of a right pyramid, of which the height is 
2 feet and the base a square on a side of 1 ft. 8 in. Ans. 10 sq. ft. 

2. Find the convex surface of a right pyramid 1 foot high, standing on 
a rectangular base whose length is 5 feet 10 inches and breadth 10 inches. 

Ans. 8 sq. ft. 128 sq. in. 

3. Find the con-vex surface of a right pyramid having the same base 
and height as a cube whose edge is 10 inches. Ans. 223.6 sq. in. 

4. Find the weight of a granite pyramid 9 feet high, standing on a 
square base whose side is 3 feet 4 inches, 1 cubic foot of granite weighing 
165 Ibs. Ans. 2 tons, 9 cwt. 12 Ibs. 

5. Find the height of a pyramid of which the volume is 623 . 52 cu. in., 
and the base a regular hexagon on a side of 1 foot. Ans. 5 inches. 



PROBLEMS. 353 

6. The volume of a regular octahedron is 471.41 cubic feet; find the 
length of each edge. Ans. 10 feet. 

7. Find the surface of a regular tetrahedron, if the perpendicular from 
one vertex to the opposite face is 5 inches. 

8. A conical vessel is 5 inches in diameter and 6 inches deep. To what 
depth will a ball 4 inches in diameter sink in the vessel ? 

9. The ends of the frustum of a pyramid are squares whose sides are 
20 inches and 4 inches, respectively. If its altitude is 15 inches, what is 
its convex surface ? Ans. 110 sq. in. 

10. What is the volume of a frustum of a pyramid whose upper base is 
4 inches square, lower base 28 inches, and the length of the slant edges 
15 inches ? 

11. The volume of a frustum of a cone is 407 cubic inches and its thick- 
ness is W}4 inches? If the diameter of one end is 8 inches, find the diam- 
eter of the other end. O=^.] Ans. 6 inches. 

SPHERES. 

1. Find the ratio of the surface of a sphere to the surface : (i) of its 
circumscribed cylinder, (ii) of its circumscribed cube. 

2. A cube and a sphere have equal surfaces ; what is the ratio of their 
volumes ? Ans. 72 : 100, nearly. 

3. From a cubical block of rubber the largest possible rubber ball is 
<mt. What decimal of the original solid is cut away? 

4. Suppose the earth to be a perfect sphere, 8,000 miles in diameter ; to 
what height would a person have to ascend in a balloon in order to see 

2^* 
one-fourth of its surface? [Formula. h 2 where r is the radius of 

the earth, and is the part of the earth's surface visible to the observer. 

If the part of the earth visible to the observer is .-, or 1/-2L, =- 

q P P 

5. A paring an inch wide is cut from a smooth, round orange an inch 
and a half in diameter. What is its volume, if it is cut from the orange 
on a great circle of the orange ? Ans. \"^. 

6. What would be the volume of a paring cut from the earth on the 
equator? Ans. $*a a t where a is the width of the paring. 

Remark This is a remarkable fact, since the volume of the paring is independent 
of the radius of the sphere. 

7. If, when a sphere of cork floats in the water, the height of the sub- 
merged segment is # of the radius, show that the weights of equal vol- 
umes of cork and water are as 3 4 :4 4 . 

Note The weight of a floating body is equal to the weight of the liquid it displaces. 

8. A vertical cylindrical vessel whose internal diameter is 4 feet, is com- 
pletely filled with water. If a metal sphere 25 inches in diameter is laid 
upon the rim of the vessel, find what weight of water will overflow. 

Ans. 699 Ibs., nearly. 

9. A conical wine-glass 5>^V 3 inches in diameter and 4 inches deep is 
filled with water. If a metal sphere 5'/ inches in diameter is placed in the 
vessel, what fraction of the whole contents will overflow ? Ans. \. 

10. Four equal spheres are tangent to each other. What is the radius 
of a sphere tangent to each ? 

11. To what depth will a sphere of ice, three feet in diameter, sink in 
water, the specific gravity of ice being -f ? 



354 FINKEL'S SOLUTION BOOK. 

12. Find the volume removed by boring a 2-inch auger-hole through a 
6-inch globe. 

13.* What is the volume removed by chiseling a hole an inch square 
through an 8-inch globe ? 

PRISMATOIDS AND WEDGES. 

1. Find the weight of a steel wedge whose base measures 8 inches by 5 
inches, and the height of the wedge being 6 inches ; if 1 cu. in. of steel 
weighs 4 . 53 oz. ? 

2. Find the volume of a prismatoid of altitude 3.5 cm., the bases being 
rectangles whose corresponding dimensions are 3 cm. by 2 cm. and 3.5 cm. 
by 5 cm. 

3. The base of a wedge is 4 by 6, the altitude is 5, and the edge, ^, is 3. 
Find the volume. 

RINGS. 

1. Find the surface and volume of a ring, the radius of the inner cir- 
cumference being 10> inches and the diameter of the cross-section 3> 
inches. Ans. 847 sq. in. ; 741 cu. in. 

2. Find the surface and volume of a ring, the diameters of the inner 
and outer circumferences being 9.8 inches and 12.6 inches respectively. 

Ans. 154.88 sq. in. ; 54.21 cu. in. 

SIMILAR SOLIDS. 

1. The edges of two cubes are as 4:3; find the ratio of their surfaces 
and their volumes. 

2. The surfaces of two spheres are in the ratio of 25:4; find the ratio 
of their volumes. 

3. At what distance from the base must a cone, whose height is 1 foot, 
be cut by a plane parallel to the base, in order to be divided into two parts 
of equal volume ? Ans. 2 . 47 in. 

4. A right circular cone is intersected by two planes parallel to the base 
and trisecting the height. Compare the volumes of the three parts into 
which the cone is divided. Ans. 1:7:19. 



EXAMINATION T^STS. 



ARITHMETIC. 

1. How do you divide one fraction by another? Why is the fraction 
thus divided ? 

2. Divide four million and four millionths by one ten - thousandth. 
Write the answer in figures and words. 

3. If a liter of air weighs 1 . 273 gr., what is the weight in kilos., if the 
air is in a room which contains 78 cu. m. ? 

4. The base of a cylinder is 12 inches in diameter and its altitude is 25 
inches. Required the solid contents. 

5. The edge of a cube is 6 inches ; what is the length of the diagonal 
of the cube ? 

6. A broker bought stock at 4% discount, and sold it at 5% premium^ 
and gained $450. How many shares did he purchase? 




PROBLEMS. 



355 



7. A ships 500 tons of cheese, to be sold at 9^ cents a Ib. He pays his 
agent 3% for selling ; the proceeds are to be invested in sugar, after a com- 
mission of 2% is deducted for buying. Required the entire commission. 

8. Upon what value are dividends declared? Brokerage estimated? 
Usual rate of brokerage ? 

9. What is the face of a note dated July 5, 1881, and payable in 4 months 
to produce $811, when discounted at 9%? 

10. Upon what principle is the United States rule for partial payments 
based? The Mercantile rule? How does compound interest differ from 
annual interest? Ohio State List, 1884. 



For the benefit of students preparing for county or state examinations, 
we write out the answers to the above questions as a specimen of how the 
examination paper ought to be prepared : 



SUBJECT : Arithmetic. 

Name of Applicant. 



(a) Invert the terms of the divisor and then multiply the 
numerators of the fractions together for the numerator of the 
quotient and the denominators together for the denominator of 
the quotient. 

(6) The fraction is thus divided because inverting the terms 
of the divisor gives the number of times the divisor is con- 
tained in 1, as is shown by analysis. The number of times then 
it is contained in any other number is obtained by multiplying 
this number by the number of times the divisor is contained 
in 1. 



II. 



1. 



Four million and four millionths=4000000. 000004= 
40000000000004 

1000000 

1 
One ten-thousandth = .0001 = 



3. 



10000 

40000000000004 1 40000000000004 10000 

X i 



1000000 ' 10000 
40000000000004 



1000000 
4 



I 



100 



-=400000000000^=400000000000 . 04= 



III. .*. The quotient is four hundred billion and four huudredths. 



II. 



III. 



1. 1 cu. m.=1000 1. 

2. 78 cu. m. =78X1000 1=78000 1. 

3. 1.273 g.=the weight of 1 1. of air, and 

4. 99294 g. =78000X1 -273 g.= weight of 78000 1. 

5. 1000 g.=l kilo. 

6. 99294 g. =99294 g. -=-1000=99. 294 kilos. 

/. The weight of 78 cu. m. of air weighs 99.294 kilos. 



356 



FINKEL'S SOLUTION BOOK. 

Arithmetic Continued. 



12 in.=the diameter of the cylinder, and 
25 in.=the altitude. Then 



II. \ 3. 
4. 



III. 



X 7rl28 = 367r s q- in., the area of the base of the cylinder. 
25X36^=900* C u. in.=900X3. 141592X1 cu. in.= 
2827.4328 cu. in., the volume of the cylinder. 
/. The volume of the cylinder is 2827 .4328 cu. in. 



1. 6 in.=the length of the edge of the cube. 

2. 36 sq. in.-|-36 sq. in. =72 sq. in. = the area of the square 

described on the diagonal of one of the equal faces, 
which is the sum of the areas of the squares de- 
scribed on two equal edges. 

** ' 3. 72 sq. in.-f-36 sq. in. =108 sq. in.=area of square de- 
scribed on the diagonal of the cube, which equals the 
sum of the areas described on the three edges. 
4. 6V3ln.=vT08Xl in. =10. 392+ in., the length of the diag- 
onal of cube. 
III. /. 6V3ln.=10.392+ in.=length of diagonal of cube. 



II. 



III. 



1. 100%= par value of stock. 

2. 4%= discount. 

3. 96% = 100% 4%= market value, or cost of stock. 

4. 5% = premium. 

5. 105% = 100% -f5%= selling price of stock. 

6. 9% = 105% 96%=gain. 

7. $450=gain. 

8. /.9% =$450. 

9. 1%=$50, and 

10. 100%=$5000=par value of stock. 

11. $100=par value of one share, usually. Then 
1 12. $5000=par value of $5000-r-$100, or 5 shares. 

.'. He purchased 5 shares. 



II. 



10 



II. 



9j4 cents=selling price of one Ib. 
$47.50=500X$0.09^=selling price of one Ib. 
fl. 100% =$17. 50. 
1% =$0.475. 

2%=2X$0.475=$0.95=commission for selling 
the cheese. 
$47.50 $0 95=$46.55=proceeds, or the amount to be 

invested in sugar. 
100%=cost of sugar. 

3%=commission on sugar. 
103%=total cost of sugar. 
$46.55=total cost of sugar, 
flv .'. 103% =$46. 55. 
j 2. l%= T fo of $46.55=$0.45. 

1 3. 100%=100X$0.45=$45=cost of sugar. 
14. 2%=2X$0.45=$0.90=commission on sugar. 

, .'. $0 . 95+$0 . 90=$1 . 85=total commission. 
$1.85=entire commission. 



PROBLEMS. 
Arithmetic Concluded. 



357 



Dividends are declared : 

(a) Upon the par value. 

(b) Brokerage is reckoned upon the selling price or purchas- 
ing price of bonds in Commission and Brokerage, but in Stock 
Investments it is reckoned on the par value. 

(c) The usual rate of brokerage is % on the par value of the 
stock, either for a purchase or a sale. 



100%=face of note. 
3^% ^discount for 126 da. 
)6 % = proceeds. 
$811=proceeds. 



1. 100%=face of note. 1881 7 

2. 3A<^=discount for 126 da. 4 
3. 

II. -I 4. 
5. 
6. 
7. 
III. " /. The face of the note must be $837.372. 



when dated. 



188111 5-8 when due. 



1% =$8. 37372, and 

=$837.372, the face of the note. 



10 



(a) Upon the principle that payments be applied first to the 
discharge of interest due, the balance, if any, toward paying the 
principal and interest. Interest or payment must in no case 
draw interest. 

(b) Upon the principle that partial payments shall draw inter- 
est from time of payment until date of settlement. 

(c) Compound Interest increases in a geometrical ratio, and 
Annual Interest in an arithmetical ratio. 



1. A and B together have $9,500. Two-thirds of A's money equals f of 
B's. How much money has each ? 

2. A owes a sum equal to f of his yearly income. By saving T V of his 
income annually for 5 years, he can pay his debt and have $1,200 left. What 
is his yearly income? 

3. Smith and Jones can do a piece of work in 12 days. If Smith can do 
only f as much as Jones, how long will it take each of them to do the work ? 

4. I am offered 6% stock at 84, and 5% stock at 72. Which investment 
is preferable, and how much ? 

5. If in selling cloth f of the gain is equal to ^ of the selling price, for 
how much will 3^ yards sell that cost $5 per yard ? 

6. The frustum of a cone is 10 feet in diameter at the bottom and 8 feet 
at the top, with a slant height of 12 feet. What is the height of the cone 
from which the frustum is cut ? 

7. A, B and C ate eight pies. If they ate equal shares and A and B furnish 
the pies, and C pays 16 cents for his share, how should A and B divide the 
money? 

8. Which is the heavier, and how much, an ounce of lead or an ounce 
of gold? Pickaway County List, f8oo. 



1. Define bonds, coupons, exchange, tariff. 

2. A field of 12 acres and 30 perches yields 255 bu. 2 qts of wheat; how 
much will a field of 15 acres and 10 perches yield at the same rate? 

3. Find value of 11% of $180 + 22% of $160 -f 92% of $63. 



358 FINKEL'S SOLUTION BOOK. 

4. A piano was sold for $297, at a gain of 35% ; what would have been 
the % of gain if it had been sold for $300 ? 

5. A dealer imported 120 dozen champagne, invoiced at $23 a dozen,, 
breakage 12% ; what was the duty at 22% ? 

6. I rent a house for $300 per year, the rent to be paid monthly in ad- 
vance ; what amount of cash at the beginning of the year will pay one 
year's rent ? . 

7. The rafters of a house are 20 feet long, the width of the gable is 30= 
feet, the rafters project two feet ; what is the height of the gable ? 

8. What the convex surface of the frustum of a cone whose slant height 
is 6 feet, the diameter of its lower base 5 feet, and of its upper 4 feet? 

9. To be analyzed : If for every cow a farmer keeps, he allows \ acre for 
pasture, and f of an acre for corn, how many cows can he keep on 39 acres ? 

10. How much can I afford to give for 6's of '81 so that I may realize 8% 
per annum, gold'being at a premium of 15? Hancock County List. 

1. What is the surface of a parallelepiped, 8 feet long, 4 feet wide, and 
2 feet high? 

2. A starts on a journey at the rate of 3 miles per hour ; 6 hours after- 
wards B starts after him at the rate of 4 miles per hour. How far will B 
travel before he overtakes A? 

3. The time since noon is ^ of the time to 4 o'clock P. M. ; what is the 
time? 

4. A man having oranges at 4 cents each, and apples at 2 for 1 cent, 
gained 20% by selling 5 dozen for $2.04 ; how many of each did he sell ? 

5. The first term of a geometric series is 3, the third term 507 ; find the 
ratio. 

6. A merchant sold a quantity of goods at a gain of 20%. If, however, 
he had purchased the goods for $60 less, his gain would have been 25%. 
What did the goods cost ? 

7. There is a park 400 feet square ; a walk 3 feet wide is made in it, along 
the edges, how many square yards would such a walk contain ? 

8. A man sold wheat, commission 3% and invested the proceeds in corn, 
commission 2% his whole commission, $250 ; for how much did the wheat 
sell and what was the value of the corn ? Licking County List. 

1. A man had 43f yards of carpeting, costing $26^ ; he sold f of the 
pieces gaining %\ on each yard sold. How much did he receive for it? 

2. From the product of f and -f^ subtract the difference of their 
squares. 

3. How many acres in a field whose length is 40 rods and diagonal 50 
rods? 

4. How many trees will be required to plant the above by placing them 
1 rod apart ? By 2 rods apart ? 

5. Bought, a lot of glass; lost 15% by breakage. At what % above cost 
must I sell the remainder to clear 20% on the whole ? 

6. After spending 25% of my money, and 25% of the remainder, I had 
left $675. How much had I at first ? 

7. How many fifths in ^? Ans. If. 

8. A box is 3 inches long, 2 inches wide, and 2 inches deep will con- 
tain how many J-inch cubes ? 

9. Change f of quart to the decimal of a bushel. 

10. A can hoe 16 rows of corn in a day, B 18, C 20, and D 24. What is 
the smallest number of rows that will keep each employed an exact num- 
ber of days ? Seneca County List. 



MENSURATION. 35 

1. (a) Define: number, integer, fraction, a common multiple, and 
the greatest common divisor of two or more numbers. 

(b) Prove (do not- merely illustrate} that to divide by a fraction 
one may multiply by the divisor inverted. 

(c) Change 74632 from a scale of 8 to a scale of 9. 

2. (a) The freezing and boiling temperatures of water are 32 and 
212, respectively, when measured by a Fahrenheit thermometer; meas- 
ured by a centigrade thermometer they are and 100, respectively; if 
a Fahrenheit thermometer records a temperature of 74 what would the 
centigrade record be at the same time? 

(b) By what per cent must 8 Fahrenheit be increased so as to- 
equal 8 centigrade? 

3. Silver weighs 10.45 times as heavy as water, while gold weighs 19.30 
times as heavy as water; find, correct to 3 decimal places, the number 
of inches in the edge of a cube of gold which is equal in weight to a cube 
of silver whose edge is 4.3 cm. Also express this weight in (Troy) grains. 

4. A 6% bond, which matures in 3 years, with interest payable annu- 
ally, is selling at 104 ; a ?% bond, which matures in H years, with interest 
payable semi-annually, is selling at 102. Which is the better investment? 
And how much better is it? 

5. A water-tank has connected with it 4 pipes; the first can fill it in 
30 min., the second in 40 min., the third can empty it in 50 min., and 
the fourth can empty it in one hour. If these pipes are so arranged that 
the third is automatically opened when the tank is precisely filled, and 
the fourth when the tank is f filled, how long will it take to just fill the 
tank if the second pipe is set running 10 minutes later than the first? 

T. 
Cornell University Scholarship Examination, 1899. 

1. A and B run a race, their rates of running being as 17 to 18. A 
runs 2 miles in 16 minutes, 48 seconds and B the whole distance in 34 
minutes. What is the distance run? 

2. The surface of the six equal faces of a cube is 1350 sq. inches. 
What is the length of the diagonal of the cube? 

3. A man bought 5% stock at 109 J, and 4J% pike stock at 107, broker- 
age in each case % ; the former cost him $200 less than the latter, but 
yielded the same income. Find the cost of the pike stock. 

4. A, B, and C start together and walk around a circle in the same 
direction. It takes A ^ hours, B f hours, C f-f- hours to walk once 
around the circle. How many times will each go around the circle before 
they will all be together at the starting point? 

5. I hold two notes, each due in two years, the aggregate face value 
of which is $1020. By discounting both at 5%, one by bank, the other 
by true discount, the proceeds will be $923. Find face of bank note. 

6. The hour and minute hands of a watch are together at 12 o'clock; 
when are they together again? 

7. How many cannon balls 12 inches in diameter can be put into a 
cubical vessel 4 feet on a side ; and how many gallons of wine will it 
contain after it is filled with the balls, allowing the balls to be hollow, 
the hollow being 6 inches in diameter, and the opening leading to it con- 
taining one cubic inch? 

8. An agent sold a house at 2% commission. He invested the pro- 
ceeds in city lots at 3% commission. His commissions amounted to $350. 
For what was the house sold? 

Ohio State List, December, 1898. 



:360 FINKEVS SOLUTION BOOK. 

1. A, B, and C can do a piece of work in 84 days; A, B, and D in 
72 days; A, C, and D in 63 days; B, C, and D in 56 days. In what 
time can each do it alone? 

2. A banker bought U. S. 4's at 128|% and U. S. 4's at 106^, 
brokerage |%. The latter cost him $1053.75 more than the former, but 
yielded him $195 more income. How much was invested in each kind 
of bonds? 

3. f of the cost of A's house increased by of the cost of his farm 
for 2 years at 5%, amounts to $4950. What was the cost of each, if f 
of the cost of the house was only f as much as of the cost of the farm? 

4. A man desiring to find the height of a tree, places a 12-foot pole 
upright 54 feet from the base of the tree; he then steps back 6 feet, and 
looks over the top of the pole at the top of the tree; his eyes are 4 
feet above the ground. How high is the tree? 

5. I have, as the net proceeds of a consignment of goods sent by 
me, $3816.48, which the consignor desires me to remit by draft at 2 months. 
If the rates of exchange are f% premium, and the rate of interest 6%, 
what will be the face of the draft? 

6. In a certain factory are employed men, women, and boys ; the boys 
receive 3 cents per hour, the women 4 cents, and the men 6 cents ; the 
boys work 8 hours per day, the women 9 hours, the men 12 hours ; the 
boys receive $5 as often as the women receive $10, and for every $10 paid 
to the women, $24 are paid to the men. How many are there of each, the 
whole number being 59? 

7. Chicago is 87 35' west. What is the standard time at Chicago 
when it is 1 P. M. at Greenwich? 

8. From the middle of the side of a square 10-acre field, I run a 
line cutting off 3$- acres. Find the length of the line. 

Ohio State List, June, 1899. 

1. How would you present to a class the subject of addition of frac- 
tions ? Take as an illustrative example, f + f + T V 

2. A reservoir is 1.50 meters wide, 2.80 meters long, and 1.25 meters 
deep. Find how many liters it contains when full, and to what height 
it would be necessary to raise it that it might contain 10 cu. meters. 

3. Reduce (a) .4685 T. to integers of lower denominations, and 
(b) 1.69408 to a common fraction in its lowest terms. 

4. The boundaries of a square and circle are each 40 feet. Which has 
the greater area and how much? 

5. Find the date of a note of $760, at 8% simple interest, which, when 
it matured December 1, 1891, amounted to $919.60. 

6. A gentleman wishes to invest in 4J% bonds, selling at 102, so as 
to provide for a permanent income of $1620. How much should he invest ? 

7. From one-tenth take one-thousandth ; multiply the remainder by 
10000 ; divide the product by one million, and write the answer in words. 

8. Bought 50 gross of buttons for 25, 10, and 5% off, and disposed 
of the lot for $35.91 at a profit of 12%. What was the list price of the 
buttons per gross? 

9. Had an article cost 10% less, the number of per cent gain would 
have been 15 more. What was the per cent gain? Give analysis. 

10. If the volume of two spheres be 100 cu. in. and 1000 cu. in. re- 
spectively. Find the ratio of their diameters to the nearest thousandth 
of an inch. 

Ohio State List, December, 1891. 



MENSURATION. 361 

PROBLEMS. 



1. What is the area of a field in the form of a parallelogram, whose 
length is 160 rods and width 75 rods? Ans. 75 A. 

2. Find the area of a triangle whose base is 72 rods and altitude 16 
rods. Ans. 3 A. 2 R. 16 P. 

3. Two trees whose heights are 40 and 80 feet respectively, stand on op- 
posite sides of a stream 30 ft. wide. How far does a squirrel leap in jumping 
from the top of the higher to the top of the lower? Ans. 50 feet. 

4. How many steps of 3 feet each does a man take in crossing diagonal 
ly, a square field that contains 20 acres? Ans. 440 steps. 

5. Find the cost of paving a court 150 feet square; a walk 10 feet around 
the whole being paved with flagstones at 54 cents a squaie yard and the 
rest at 31> cents a square yard? Ans. $939.40. 

6. What is the area of a triangle, the three sides of which are respect- 
ively 180 feet, 150 feet, and 80 feet? Ans 5935.85 sq. ft. 

7. What is the area of a trapezium, the diagonal of'which is 110 feet, and 
the perpendiculars to the diagonal are 40 feet and 60 feet respectively ? 

Ans. 5500 sq. ft. 

8. At 30 cents a bushel, find the cost of a box of oats, the box being & 
feet long, 4 feet wide and 4 feet deep. Ans. $30.85)^. 

9. Two trees stand on opposite sides of a stream 40 feet wide. The 
height of one tree is to the width of the stream as 8 is to 4, and the width of 
the stream is to the height of the other as 4 is to 5. What is the distance 
between their tops ? Ans. 50 feet. 

10. How many miles of furrow 15 in., wide, is turned in plowing a rect- 
angular field whose width is 30 rods and length 10 rods less than its diagoal ? 

Ans. 4Q% mi. 

11. The sides of a certain trapezium measure 10, 12, 14, and 16 rods 
respectively, and the diagonal, which forms a triangle with the first two 
sides, is 18 rods; what is the area? Ans. 163.796 sq. rds. 

12. Three circles, each 40 rods in diameter, touch each other externally; 
what is the area of the space inclosed between the circles ? 

Ans. 64.5 sq. rds. 

13. How many square *ncnes in one face of a cube which contains 2571353 
cubic inches? Ans. 18769 sq. in. 

14. Four ladies bought a ball of thread 3 inches in diameter; what por- 
tion of the diameter must each wind off to heve equal shares of the thread? 

First, .2743191 in. 
\ Second, .3445792 in. 
. -j Third? .4912292 in. 
1 Fourth, 1.8898815 in. 

15. A gentleman proposed to plant a vineyard of 10 A. If he pla'ces the 
vines 6 feet apart; how many more can he plant by setting them in the 
quincunx order than in the square order, allowing the plat to lie in the form 
of a square, and no vine to be set nearer its edge than 1 foot in either case? 

Ans. 1870. 

16. Find the volume generated by the revolution of a circle about a 
tangent. Ans. 2n z R z . 

17. How many feet in aboard 14 feet long and 16 inches wide at one 
end and 10 inches at the other, and 3 inches thick? Ans. 45^ feet. 

18. If I saw through^ of the diameter of a round log, what portion of 
the cut is made? Ans. .196. 



362 FINKEL'S SOLUTION BOOK. 

T9. .What is the surface of the largest cube that can be cut from a sphere 
which contains 14137.2 cu. ft.? Ans. 1800 sq. ft. 

20. Two boys are flying a kite. The string is 720 feet long. One boy 
who stood directly under the kite, was 50 feet from the other boy who held 
the string; how high was the kite? Ans. 717.8-|-feet. 

21. How many pounds of wheat in a cylindrical sack whose diameter is 
1> feet, and whose length is 1% yards? (?r:=3.1416) sns. 447.31 Ib. 

22. How large a square can be cut from a circle 50 inches in diameter? 

A ns. 35.3553391 in. 

23. How many bbl. in a tank in the form of the frustum of a pyramid, 
5 feet deep, 10 feet square at the bottom and 9 feet square at the top? 

Ans. 107.26 bbl. 

24. From a circular farm of 270 acres, a father gives to his sons equal 
circular farms, touching each other and the boundary of the farm. He 
takes for himself a circular portion in the center, equal in area to a son's 
part, and reserves the vacant tracts around his part for pasture lands and 
gives each son one of the equal spaces left along the boundary. Required 
the number of sons and the amount of pasture land each has. 

Ans. 6 sons; 8.46079 A. 

25. At each angle of a triangle being on a level plain and having sides 
respectively 40, 50, and 60 feet, stands a tower whose height equals the sum 
of the two sides including the angle. Required the length of a ladder to 
reach the top of each tower without moving at the base. 

Ans. 116.680316-fft. 

26. If the door of a room is 4 feet wide, and is opened to the angle of 90 
degrees, through what distance has the outer edge of the door passed? 

Ans. 6.2832 feet. 

27. A tinner makes two similar rectangular oil cans whose inside dimen- 
sions are as 3, 7, and 11. The first hold 8 gallons and the second being 
larger requires 4 times as much tin as the other. What are the dimensions 
of the smaller and the contents of the larger ? 

. j Dimensions of smaller 6, 14, and 22 inches. 
1ls ' ( Capacity of larger 64 gallons. 

28. An 8-inch globe is covered with gilt at 8 cents per square inch; find 
the cost. A ns. $16.08. 

29. A hollow cylinder 6 feet long, whose inner diameter is 1 inch and 
outer diameter two inches, is transformed into a hollow sphere whose outer 
diameter is twice its inner diameter; find outer diameter. Ans. 3.59 in 

30. A circular field is 360 rods in circumference; what is the diagonal of 
a square field containing the same area? Ans. 20.3 rods. 

31. What is the volume of a cylinder, whose length is 9 feet and the cir- 
cumference of whose base is 6 feet? Ans. 25.78 cu. ft. 

32. How many acres in a square field, the diagonal being 80 rods? 

Ans. 20 acres. 

33. How many cubical blocks, each edge of which is ^3 of a foot, will 
fill a box 8 feet long, 4 feet wide, and 2 feet thick. Ans. 1728 blocks. 

34. From one corner of a rectangular pyramid 6 by 8 feet, it is 19 feet 
to the apex; find the dimentions of a rectangular solid whose dimensions are 
as 2, 3, and 4, that may be equivalent in volume. Ans. 4, 9, and 8 feet. 

35.* A solid metal ball, 4 inches radius, weighs 8 Ibs.; what is the thick- 
ness of spherical shell of the same metal weighing 7% Ib., the external di- 
ameter of which is 10 inches? Ans. 1 inch. 

36. What is the difference between 25 feet square and 25 square feet? 

Ans. 600 sq.ft. 



MENSURATION. 363 

37.* Find the greatest number of trees that can be planted on a lot il 
rods square, no two trees being nearer each other than one rod? 

Ans. 152 trees. 

38.* A straight line 200 feet long, drawn from one point in the outer 
edge of a circular race track to another point in the same, just touches the 
inner edge of the track. Find the area of the track and its width. 

Ans. Area, 7r 2 ^=10000~ sq. ft.; width, indeterminate. 

39. The perimeter of a certain field in the form of an equilateral triangle 
is 360 rods; what is the area of the field ? Ans. 543.552 sq. rd. 

40. A room is 18 feet long, 16 feet wide, and 10 feet high. What length 
of rope will reach from one upper corner to the opposite upper corner and 
touch the floor? Ans. 35 3 ft. 

41. How many bushels of wheat in a box whose length is twice its width, 
and whose width is 4 times its height; diagonal being 9 feet? 

Ans. 25 bu., nearly. 

42 Find the area of a circular ring whose breadth is 2 inches and inside 
diameter 9 inches. Ans. 69.1152 sq. in. 

43 * A round stick of timber 12 feet long, 8 inches in diameter at one 
end and 16 inches at the other, is rolled along till the larger end describes a 
complete circle. Required the circumference of the circle. 

Ans. 150.83 feet. 

44. A fly traveled by the shortest possible -route from the lower corner 
to the opposite upper corner of a room 18 feet long, 12 feet wide and 10 feet 
high. Find the distance it traveled Ans. 28.42534 feet. 

45.* From the middle of one side and through the axis perpendicularly 
of a right triangular prism, sides 12 inches, I cut a hole 4 inches square. 
Find the volume removed. Ans. 138.564064 cu. in. 

46.* Two isosceles triangles have equal areas and perimeters. The base 
of one is 24 feet, and one of the equal sides of the other is 29 feet. The 
area of both is 10 times the area of a triangle whose sides are 13, 14, and 15 
feet. Find the perimeters and altitudes. 

Ans. Perimeters, 98 feet; altitudes 35 and 21 feet. 

47. A grocer at one straight cut took off a segment of a cheese which 
had ^ of the circumference, and weighed 3 pounds; what did the whole 
weigh? A ns. 33.023 lb. 

48.* A twelve inch ball is in a corner where walls and floor are at right 
angles; what must be the diameter of another ball which can touch that 
ball while both touch the same floor and the same walls? 

Ans. 3.2154 in. or 44.7846 in. 

49. What will it cost to paint a church steeple, the base of which is an 
octagon, 6 feet on each side, and whose slant height is 80 feet, at 30 cents 
per square yard? Ans. $64. 

50. A tree 48 feet high breaks off; the top strikes the level ground 24 feet 
from the bottom of the tree; find the height of the stump. Ans. 18 feet. 

51. How many acres in a square field whose diagonal is 5^ rods longer 
than one of its sides? . Ans. 160.6446 sq. rd. 

52.* Three poles of equal length are erected on a plane so that their tops 
meet, while their bases are 90 feet apart, and distance from the point where 
the poles meet to the center of the triangle below is 65 feet. What is the 
length of the poles ? Ans. 83.23 feet. 

53. A field contains 200 acres and is 5 times as iong as wide. What will 
it cost to fence it, at a dollar per rod ? Ans. $960. 

54.* What is the greatest number of plants that can be set on a circular 
piece of ground 100 feet in diameter, no two plants to be nearer each other 
than 2 feet and none nearer the circumference than 1 foot? Ans. 2173. 



364 FINKEL'S SOLUTION BOOK. 

55. The axes of an ellipse are 100 inches and 60 inches; what is the dif- 
ference in area between the ellipse and a circle having a diameter equal to 
the conjugate axis ? A ns. 600 ^=1884.96 sq. in. 

56. Find the diameter of a circle of which the altitude of its greatest in- 
scribed triangle is 25 feet. Ans. 33 J/ feet. 

57. If we cut from a cubical block enough to make each dimension 1 
inch shorter, it will lose 1657 cubic inches, what are the dimensions? 

58. Show that the area of a rhombus is one-half the rectangle formed by 
its diagonals. Noble Co. Ex. Test. 

59. The length and breadth of a rectangular field are in the ratio of 4 
to 3. How many acres in the field, if the diagonal is 100 rods ? 

60. A spherical vessel 30 inches in diameter contains in depth, 1 foot of 
water; how many gallons will it take to fill it? Holmes Co. Ex. Test. 

Ans. 39 gallons. 

61. A field is 40 rods by 80 rods. How long a line from the middle of 
one end will cut off 7) acres? Ans. 80.6 rd., nearly. 

62. A ladder 20 feet long leans against a perpendicular wall at an angle 
of 30. How far is its middle point from the bottom of the wall? 

Ans. 10 feet. 

63. Four towers, A 125 feet high, B 75 feet, C 160 feet, and D 65 feet, 
stand on the same plane. B due south and 40 rods from A; C east of B 
and D south of C. The distance from A to C plus the distance from C to B 
is half a mile, and the distance from D to B is 82}^ yd. farther than the dis- 
tance from C to D. What length of line is required to connect the tops of 
AandD? Ans. 240+rus. 

64. Find the volume of the largest square pyramid that can be cut from 
a cone 9 feet in diameter and 20 feet high? Ans. 270 cu. ft. 

65. A rectangular lawn 60yd. long and 40 yd. wide has a walk 6ft. wide 
around it and paths of the same width through it, joining the points of the 
opposite sides. Find in square yards the area of one of the four plats in- 
closed by paths. Ans. 459 sq. yd. 

66. Which has the greater surface, a cube whose volume is 13.824 cu. ft., 
or a rectangular solid of equal volume whose length is twice its width, and 
its width twice its height? Ans. Rect. 576 sq. ft, more. 

67. The volume of a rectangular tin can is 3 cu. ft. 1053 cu. in.; its di- 
mensions are in the proportion of 11, 7, and 3. Find the area of tin in the 
can. Ans. 16% sq. ft. 

68. A conical well has a bottom diameter of 28 ft. 3 in., top diameter 
56 ft. 6 in., and depth 23 ft. 1.2 in. Find its capacity in barrels. 

Ans. 8023 bbl. 

69. A cylindrical vessel 1 foot deep and 8 inches in diameter was |f full 
of water; after a ball was dropped into the vessel it was full. Find the di- 
ameter of the ball. Ans. 6 inches. 

70. Two logs whose diameters are 6 feet lie side by side. What is the di- 
ameter of a third log placed in the crevice on top of the two, if the pile is 9 
feet high?. Ans. 4 ft. 

71. Circles 6 and 10 feet in diameter touch each other; if perpendiculars 
from the center are let fall to the line tangent to both circles, how far apart 
will they be? Ans. 7.756 ft. 

72. What are the linear dimensions of a rectangular box whose capacity 
is 65910 cubic feet; the length, breadth, and depth being to each other as 
5, 3, and 2? Ans. 65, 39, and 26 ft. 

73. The perimeter of a piece of land in the form of an equilateral trian- 
gle is 624 rods; what is the area? Ans. 117 A. 13 31 P. 



MENSURATION. 365 

74. Four logs 4 feet in diameter lay side by side and touch each other; 
on these and in the crevices lay three logs 3 feet in diameter; on these three 
and in the crevices lay two logs 2 feet in diameter; what is the diame- 
ter of a log that will lay on the top of the pile touching each of the logs 2 
feet in diameter and the middle one of the logs 3 feet in diameter? 

Ans. 

75. What will it cost to gild a segment of a sphere whose diameter is 6 
inches; the altitude of the segment being 2 inches, at 5 ^ per square inch? 

76. A grocer cut off the segment of a cheese, and found it took of the 
circumference. What is the weight of the whole cheese, if the segment 
weighed \% Ibs ? Ans. 52.0228+lbs. 

77. Two ladders are standing in the street 20 feet apart. They are in- 
clined equally toward each other at the top, forming an angle of 45. Find, 
by arithmetic, the length of the ladders? Ans. 26.13 ft. 

Union Co. Ex. List. 

78. Two trees stand on opposite sides of a stream 120 feet wide; the height 
of one tree is to the width of the stream as 5 is to 4, and the width of the 
stream is to the height of the other as 5 is to 4; what is the distance between 
their tops? Ans. 131.58 ft. 

79. How many gallons of water will fill a circular cistern 6 feet deep 
and 4 feet in diameter? Ans. 564.0162 gal. 

80. A cube of silver, whose diagonal is 6 inches, was evenly plated 
with gold; if 4 cubic inches of gold were used, how thick was the plating? 

Ans. iV in. 

81. Required the distance between the lower corner and the opposite 
upper corner of a room 60 feet long, 32 feet wide, and 51 feet high ? 

Ans 85 ft. 

82. How deep must be a rectangular box whose base inside is 4 inches 
by 4 inches to hold a quart, dry measure? Ans. 4.2 cu in 

83. A fly is in the center of the floor of a room 30 feet long, 20 feet 
wide, and 12 feet high. How far will it travel by the shortest path to one 
of the upper corners of the ceiling? Ans. v/709-j-ft 

84. A corn crib 25 feet long holds 125 bushels. How many bushels will 
one of like shape and 35 feet long hold ? 

85. Let a cube be inscribed in a sphere, a second sphere in this cube, a 
second cube in this sphere, and so on; find the diameter of the 7th sphere, 
if thatof the first is 27 inches. (2). What is the volume of all the spheres 
so inscribed including the first? Ans. . 

86. The area of a rectangular building lot is 720 sq. ft ; its sides are as 
4 to 5; what will it cost to excavate the earth 7 feet deep at 36^ per cubic 
yard? Ans. $67.20. 

87. A owns ^ and B the remainder of a field 60 rods long and 30 rods 
wide at one end and 20 rods wide at the other end, both ends being parallel 
to the same side of the neid They propose to lay out through it, parallel 
with the ends, a road one rod wide leaving A's % of the remainder at the 
wide end and B's % at the narrow end of the field. Required the location 
and area of the road. Ans . 

88. The diameter of a circular field is 240 rods. How much grass will 
be left after 7 horses have eaten all they can reach, the ropes which are al- 
lowed them being of equal lengths and attached to posts so located that each 
can touch his neighbor's territory and none can reach beyond the boundary 
of the field? Ans. 62.831853 A. 

89. What is the diameter of a circle inclosing three equal tangent circles, 
if the area inclosed by the three equal circles is ] acre? Ans. . 



366 FINKEIv'S SOLUTION BOOK. 

90. What is the diameter of a circle inclosing four equal tangent circles 
each being tangent to the the required circle, if the area inclosed by the 
four equal circles is I acre? Ans. /?=4j/[5(4 7r )](v/2+l)-f-(4 TT). 

91. What is the greatest number of stakes that can be driven one foot 
apart on a rectangular lot whose length is 30 feet and width 20 feet? 

A ns. . 

92. What is the greatest number of inch balls that can be put in a box 15 
inches long, 9 inches wide, and 6 inches high? Ans. . 

93. A conical vessel 6 inches in diameter and 10 inches deep is full of 
water. A heavy ball 8 inches in diameter, is put into the vessel; how much 
water will flow out? Ans . 

94. How far above the surface of the earth would a person have to ascend 
in order that % of its surface would be visible? Ans. 8000 mi. 

95. Where must a frustum of a cone be sawed in two parts, to have 
equal solidities, if the frustum is 10 feet long, 2 feet in diameter at one end, 
and 6 feet at the other? Ans . 

96. At the three corners of a rectangular field 50 feet long and 40 feet 
wide, stands three trees whose heights are 60, 80, and 70 feet. Locate 
the point where a ladder must be placed so that without moving it at the 
base it will touch the tops of the three trees, and find the length of the lad- 
der. What must be the height of a tree at the fourth corner so that the 
same ladder will reach the top, the foot of the ladder not being moved? 

Ans. . 

97. A horse is tied to a corner of a barn 50 feet long and 30 feet wide; 
what is the area of the surface over which the horse can graze, if the rope 
is 80 feet long? Ans. . 

98 How many cubic feet in a stone 32 feet high, whose lower base is a 
rectangle, 10 feet by 4 feet and the upper base 8 feet by 1% feet? 

Ans. 805^ cu. ft. 

99. To what height above the ground would a platform, 10 feet by 6 
feet, have to be elevated so that 720 sq. ft. of surface would be invisible to a 
man standing at the center of the platform, the man being 5 feet high?. 

100. Required the side of the least equilateral triangle that will cir- 
cumscribe seven circles, each 20 inches in diameter. Ans. 89 28203 in. 

101. Required the sides of the least right triangle that will circumscribe 
seven circles each 20 inches in diameter. Ans 123.9320 in. and 

107.3205 in. 

102. How long a ladder will be required to reach a window 40 feet from 
the ground, if the distance of the foot of the ladder from the wall is f of the 
length of the ladder. Ans. 50 ft. 

103. A circular park is crossed by a straight path cutting off ^ of the 
circumference; the part cut off contains 10 acres Find the diameter of the 
park. Ans. 150 rd., nearly 

104. Find the length of the minute-hand of a clock, whose extreme point 
moves 5 ft. 5.9736 in., in 1 da. 18 hr. ? Ans. \ in. 

105. A, B, and C, own a triangular tract of land. Their houses are located 
at the vertices of the triangle; where must they locate a well to be used in 
common so that the distance irom the houses to the well will be the same, 
the distance from A to B being 120 rods, from B to C 90 rods and A to C 
80 rods. Ans. . 

106. A horse is tethered from one corner of an equilateral triangular 
building whose sides are 100 feet, by a rope 175 feet long. Over what area 
can he graze? Ans. 90021.109181 sq. ft. 

107. Find the area of the triangle formed by joining the centers of the 
squares constructed on the sides of an equilateral triangle, whose sides are 
20 feet ? A ns. 



GEOMETRY. 



367 



GEOMETRY. 



I. DEFINITIONS. 

i. Geometry is that branch of mathematics which deduces 
the properties of figures in space from their defining conditions, 
by means of assumed properties of space. Century Dictionary. 

ri 3 Plane 



j. Platonic 
Geometry - 



2. Geometry 



1. Metrical 
Geometry 



1 2 . Pure 

Geometry -| 

1 8 

2 2 . Conic Secions 



Geometry 

Solid 

Geometry 



.3 8 .Trigon'try- 
2 i . Analytical Geometry 



2. Descriptive Geometry, or Projective Geometry 



1 3 . Plane 

Trig. 
2 8 . Analytical 

Trig. 
3 3 . Spherical 

Trig. 



3. Metrical Geometry is that branch of Geometry which 
treats of the length of lines and the magnitudes of angles, areas, 
and solids. 

The fundamental operation of metrical relations is MEASUREMENT. 
The geometry of Euclid and the Ancients is almost entirely metrical. The 
theorem, The square described on the hypotenuse of a right-angled triangle 
is equal to the sum of the squares described on the other two sides, is a theo- 
rem of metrical geometry. 

4. Descriptive Geometry, also called Projective Geom- 
etry, Modern Synthetic Geometry, and Geometry of Position, is 
that branch of Geometry which treats of the positions, the direc- 
tions, and intersections of lines, the loci of points, and the nature 
and character of curves and surfaces. 

The fundamental operations of Descriptive Geometry are PROJECTION 
and SECTION. Many of the theorems of Descriptive Geometry are very 
old, dating as far back as the time of Euclid, but the theories and methods 
which make of these theorems a homogeneous and harmoneous whole is 
modern having been discovered or perfected by mathematicians of an age 
nearer our own, such as Monge, Carnot, Brianchon, Poncelet, Moebius, 
Steiner, Chasles, von Staudt, etc., whose works were published in the 
earlier half of the present century. Of the synonymous terms I have used 
to designate this geometry of which I am speaking, the term, Modern Syn- 
thetic Geometry is the most comprehensive. Descriptive Geometry was 
invented by Gaspard Monge (1746-1818) in 1794 and at that time embraced 
only the theory of making projections of any accurately defined figure such 
that from these projections can be deduced, not only the projective proper- 
ties of the figure, but also its metrical properties. Now this term is used 
to designate the entire theory and development of geometry as embraced 
in the above definition. 



368 



FINKEL'S SOLUTION BOOK. 



The problem, To draw a third straight line through the inaccessable 
point of intersection of two (converging) straight lines, is both metrical 
and descriptive, that is to say, the required line may be found either by 
metrical or descriptive geometry, but the method by Descriptive Geometry 
is far the simpler. 

The following are the solutions by both methods : 



METRICAL. 




I. Given the two converg- 
ing lines AB and CD which 
do not intersect in an acces- 
sable point. 

II. Required to draw a 
third line through the inac- 
cessable point K. 



1. Draw the transversal 
LM, intersecting AB and 
CD in E and F respec- 
tively. 

2 . Draw A^parallel to EF 

and intersecting AB and 
CD in G and H respec- 
tively. 

3 . Divide EF in any ratio , 

say 1:2, and let Q be the 
point of division. 

4. Divide GH in the same 
ratio and let R be the 
point of division. 

5. The line through QR 
is the line required. 



\J 

I 



i 



fl. Suppose the line join- 
ing the inaccessable point 
K and the point Q to in- 
tersect NP in R', if not 
inR. 



DESCRIPTIVE. 




1. Choose some point P out- 
side the two given straight 
lines AB and CD. 

2. Pass through this point 
any number of transversals, 
as FP t HP, KP. 

3. Draw the diagonals FG r 
HE, HI, and KG. 

4. The points of intersection 
L and M lie upon the line 
which passes through the 
point of intersection of AB 
and CD. 

The proof of this follows from 
the important harmonic prop- 
erties of a quadrangle. 






t 




^5r c/ u 



GEOMETRY. 



369 



, 



Then, from similar tri- 
angles, KR':KQ=R'G: 
QE. 

Also, KR' : KQ=RH\ 



.-. R'G:QE=R'H:QE. 

But, Q:QF=1:2. By 

Hyph. 



9, 



=l:2. By 

Const. 

.'.RG:RH=R'G:R'H. 

.'.RG=R'G and the 
point ^'coincides with./?. 



Many of the properties of the Conic Sections which are estab- 
lished with great labor and difficulty by Analytical Geometry are 
easily and elegantly proved by Descriptive Geometry. Descriptive 
Geometry stands among the first of the branches of pure mathe- 
matics in point of interest and simplicity of its methods. The 
best works on this subject are Luigi Cremona's Elements of Pro- 
jective Geometry, translated by Charles Leudesdorf, and Theodore 
Reye's Lectures on Geometry of Position, Part I., translated by 
Thomas F. Holgate. 

II. ON GEOMETRICAL REASONING. 

5. On Geometrical Reasoning. We are accustomed to 
speak of mathematical reasoning as being above all other, in 
accuracy and soundness. This is not correct, if we mean by 
reasoning the comparing together of different ideas and pro- 
ducing other ideas from the comparison ; for, in this view, mathe- 
matical reasonings and all other reasonings correspond precisely. 
The nature of establishing mathematical truths, however, is 
totally different from that of establishing a truth in history, 
political economy, or metaphysics, and the difference is this, viz., 
instead of showing the contrary of the proposition asserted to 
be only improbable, it proves it at once to be absurd and impos- 
sible. For example, suppose one were to ask for the proof of 
the assassination of Caesar, what would be the method of proof? 
No one living to-day is absolutely certain that Caesar was assas- 
sinated, and, in order to establish this truth, we refer to the testi- 
mony of historians, men of credit, who lived and wrote their 
accounts in the very time of which they write; the statements 
of these historians have been received by succeeding ages as 
true ; and succeeding historians have backed their accounts by a 
mass of circumstantial evidence which makes it the most improb- 
able thing in the world that the account or any particular part of 
it is false. In this way we have proved that the truth of the 



370 FINKEL'S SOLUTION BOOK. 

statement rests on a very high degree of probability, though it 
does not rise to absolute certainty. 

"In mathematics, the case is wholly different. It is true that 
the facts asserted in these sciences are of a nature totally distinct 
from those of history; so much so, that a comparison of the 
evidence of the two may almost excite a smile. But if it be 
remembered that acute reasoners, in every branch of learning, 
have acknowledged the use, we might almost say the necessity 
of a mathematical education, it must be admitted that the points 
of connection between these pursuits and others are worth attend- 
ing to. They are the more so, because there is a mistake into 
which several have fallen, and have deceived others, and per- 
haps themselves, by clothing some false reasoning in what they 
called a mathematical dress, imagining that, by the application 
of mathematical symbols to their subject they secured mathe- 
matical argument. This could not have happened if they had 
possessed a knowledge of the bounds within which the empire 
of mathematics is contained. That empire is sufficiently wide, 
and might have been better known, had the time which has been 
wasted in aggressions upon the domains of others, been spent 
in exploring the immense tracts which are yet untrodden/'* In 
establishing a mathematical truth, instead of referring to authority, 
we continually refer our statements to more and more evident 
statements, until at last we come either to definitions or to state- 
ments so evidently true, that to deny them would prove the un- 
soundness of him who makes the denial. 

Geometry must have recourse to the outside world for its 
first notions and premises, and is, therefore, a natural science. 

Yet there is a great difference, between it and the other natural 
sciences. For example, contrast Geometry and Chemistry. Both 
derive their constructive materials from sense-perception ; but 
while Geometry is compelled to draw only its first results from 
observation and is then in a position to move forward deductively 
to other results without being under the necessity of making 
fresh observations, Chemistry, on the other hand, is still com- 
pelled to make observations and to have recourse to nature. 



III. ON THE ADVANTAGES DERIVED FROM THE 
STUDY OF GEOMETRY, AND MATHEMATICS 
IN GENERAL. 

6. On the Advantages derived from the Study of 
Geometry and Mathematics in General. The story 
is told of Abraham Lincoln that before he began the study of 
law, he worked through Euclid in order to give his mind that 
training in logical thinking so necessary to a successful lawyer; 

*DeMorgan, Study of Mathematics. 



GEOMETRY. 371 

and his great success as a lawyer and statesman is largely to be 
attributed to the discipline he thus received. 

There should be no conflict between the sciences and the 
classics. A student taking a college course should give his time 
to study in both. The study of language enables a person to 
express his thoughts accurately and clearly while the study of 
the sciences provides him with thoughts worthy of expression. 
How far each of these two great departments should be pursued 
by the student, must be determined by the student himself. But 
certainly neither should be pursued exclusively. Yet if one 
were to pursue one or the other of these two great departments 
of knowledge exclusively, I heartily agree with Professor Earnst 
Mach who says, "Here I may count upon assent when I say 
that mathematics and the natural sciences pursued alone as 
means of instruction yield a richer education in matter and 
form a more general education, an education better adapted to 
the needs and spirit of the time, than the philological 
branches pursued alone would yield."* As to mathematics, 
"It is admitted by all that a finished or even a competent 
reasoner is not the work of nature alone ; the experience of every 
day makes it evident that education develops faculties which 
would otherwise never have manifested their existence. It is, 
therefore, as necessary to learn to reason before we can expect 
to be able to reason, as it is to learn to swim or fence, in order 
to attain either of these arts. Now, something must'be reasoned 
upon, it matters not much what it is, provided it can be reasoned 
upon with certainty. The properties of mind or matter, or the 
study of languages, mathematics, or natural history, may be 
chosen for this purpose. Now, of all these, it is desirable to 
choose the one which admits of the reasoning being verified, that 
is, in which we can find out by other means, such as measurement 
and ocular demonstrations of all sorts, whether the results are 
true or not. . . . Now the mathematics are peculiarly well 
adapted for this purpose, on the following grounds : 

i. Every term is distinctly explained, and has but one mean- 
ing, and it is rarely that two words are employed to mean the 
same thing. 

2. The first principles are self-evident, and, though derived 
from observation, do not require more of it than has been made 
by children in general. 

3. The demonstration is strictly logical, taking nothing for 
granted except the self-evident first principles, resting nothing 
upon probability, and entirely independent of authority or 
opinion. 

4. When the conclusion is attained by reasoning, its truth 
or falsehood can be ascertained, in -geometry by actual measure- 

*See Professor Mach's Popular Scientific Lectures, "On Instruction in the Classics 
and Sciences." Also Grant Allen's Article in the Oct. No. of the Cosmopolitan for 1897. 



372 FINKEL'S SOLUTION BOOK. 

ment, in algebra by common arithmatical calculation. This gives 
confidence, and is absolutely necessary, if, as was said before, 
reason is not to be instructor, but pupil. 

5. There are no words whose meanings are so much alike 
that the ideas which they stand for may be confounded. Be- 
tween the meanings of terms there is no distinction, except ab- 
solute distinction, and all adjectives and adverbs expressing dif- 
ference of degree are avoided. Thus it may be necessary to say, 
"A is greater than B ;" but it is entirely unimportant whether 
A is very little greater than B or very much greater than B. 
Any proposition which includes the foregoing assertion will 
prove its conclusions generally, that is, for all cases in which A 
is greater than B, whether the difference be great or little. . . . 

These are the principal grounds on which, in our opinion, the 
utility of mathematical studies may be shown to rest, as a dis- 
cipline for the reasoning powers. But the habits of mind which 
these studies have a tendency to form are valuable in the highest 
degree. The most important of all is the power of concen- 
trating the ideas which a successful study of them increases 
where it did exist and creates where it did not. A difficult 
position, or a new method of passing from one proposition to 
another, arrests all the attention and forces the united faculties 
to use their utmost exertions. The habit of mind thus formed 
soon extends itself to other pursuits, and is beneficially felt in 
all the business of life. 

"As a key to the attainment of other sciences, the use of the 
mathematics is too well known to make it necessary that we 
should dwell on this topic. In fact, there is not in this country 
any disposition to undervalue them as regards the utility of their 
applications. But though they are now generally considered as 
a part, and a necessary one, of a liberal education, the views 
which are still taken of them as a part of education by a large 
proportion of the community are still very confined."* 

The advantages derived from a study of geometry, though 
very great, are only part of those to be derived from a thorough 
course of study in mathematics. The eminent mathematician 
Cayley, "the central luminary, the Darwin of the English School 
of Mathematicians," as Sylvester calls him, said once that if 
he had to make a defence of mathematics he would do it in the 
manner in which Socrates, in Plato's "Republic" defended jus- 
tice. Justice, according to the Greek sage, was a thing desir- 
able, in itself and for its own sake, quite irrespective of the 
worldly advantages which might accompany a life of virtue 
and justice. So just for the sake of learning the beauties and 
the purest truths which mathematics, the oldest and the noblest, 
the grandest and the most profound of all sciences, represents, 

*DeMorgan, The Study of Mathematics. 



GEOMETRY. 373 

would it be worth while to make ourselves acquainted with its 
uses as an educational medium and the application it finds in 
other sciences? Sylvester says, "The world of ideas which 
mathematics discloses or illuminates, the contemplation of divine 
beauty and order which it induces, the harmonious connection 
of its parts, the infinite hierarchy and absolute evidence of truths 
with which mathematical science is concerned, these, and such 
like, are the surest grounds of its title to human regard." 
Sylvester, twenty-five years ago called the attention of the Royal 
Society to the parallelism between the mathematical and musical 
ethos : music being the mathematics of the senses, mathematics 
the music of reason; the soul of each the same. Music the 
dream, mathematics the working life ; each to receive its con- 
summation from the other, when the human intelligence elevated 
to its perfect type, shall shine forth glorified in some future 
Beethoven-Gauss. 

There is surely something in the beauty of the truths them- 
selves. They enrich us by our mere contemplation of them. 
What a charm and what a wealth of delight and self-content- 
ment does the finding of mathematical truths afford. In this 
science, of which geometry is one, out of a few postulates and 
germinating truths, the mind of man can gradually unfold a 
system of new and beautiful truths never dreamt of before. 
Locke says, "The mathematician from very plain and easy be- 
ginnings, by gentle degrees, and a continued chain of reason- 
ings, proceed to the discovery and demonstration of truths that 
appear at first sight beyond human capacity." Because mathe- 
matics is a science of pure reason and rigorous logic a mathe- 
matician may forget all the preceding propositions of his science 
and still be able to guide himself with the utmost confidence 
through the labyrinth of ideas and reach its exit, if he only 
keeps clearly before him the ends of the threads of thought. 

"It is due to the peculiarity of Mathematics, which is a chain 
of inseparable reasonings, that one part of it can hardly be 
studied to the exclusion of the others; that in order to under- 
stand the whole, only hard and persistent work, the greatest 
perseverance and the greatest caution, in which all our mental 
powers and capabilities have to be brought into play, can lead 
us to the great victory of the mind and enable us to comprehend 
and see the beauties of pure truths which this magnificent branch 
of Science represents. To all these peculiarities is due the fact 
that only a limited number of people are capable of appreciating 
the beauties of this oldest of all sciences." No fault has ever 
been found with Mathematics by the true student. He who 
has the courage to study diligently in any line of work, can 
obtain the same results when studying Mathematics with the 
same diligence and care. As the drill will not penetrate the 
granite unless kept to the work hour after hour, so the mind 



374 FINKEL'S SOLUTION BOOK. 

will not penetrate the secrets of Mathematics unless held long 
and vigorously to the work. As the sun's rays burn only when 
concentrated, so the mind achieves mastery in Mathematics and 
indeed in every branch of knowledge only when its possessor 
hurls all his forces upon it. Mathematics, like all the other 
sciences, opens its door to those only who knock long and hard. 
No more damaging evidence can. be adduced to prove the weak- 
ness of character than for one to have aversion to mathematics ; 
for whether one wishes so or not, it is nevertheless true, that 
to have aversion for mathematics means to have aversion to 
accurate, painstaking, and persistent hard study and to have 
aversion to hard study is to fail to secure a liberal education, 
and thus fail to compete in that fierce and vigorous struggle for 
the highest and the truest and the best in life which only the 
strong can hope to secure. 

But we do not judge a painting by the number of its admirers. 
It is as a rule the lowest kind of art which attracts the largest 
number of admirers. 

In this practical world, in this world of hard struggle for life, 
where the guiding principle is "swim who can and those who 
can't may drown," it may not, perhaps, be admissable to judge 
of the value of a science by its inherent beauty, but rather by 
the share it contributes to the education of our mental faculties 
and by the applications it finds in the useful arts and sciences 
and thus in what measure it contributes to the civilization of 
the world. He who reads history with some critical judgment 
cannot fail to notice that the degree of civilization of a country 
is closely connected with the standard of Mathematics in that 
country, and this fact is attested by the fierce bidding for the 
best mathematicians in the world by such countries as France, 
Russia, and Prussia during the latter part of the last century. 
Prof. H. J. Stephen Smith, of Oxford, says, "I should not wish 
to use words which may seem to reach too far, but I often find 
the conviction ,forced upon me that the increase of mathematical 
knowledge is a necessary condition for the advancement of 
science, and if so, a no less necessary condition for the improve- 
ment of mankind. I could not augur well for the enduring in- 
tellectual strength of any nation of men, whose education was 
not based on solid foundation of mathematical learning and 
whose scientific conception, or in other words, whose notions 
of the world and of things in it, were not braced and girt 
together with a strong framework of mathematical reasoning." 

Fourier, one of the greatest mathematicians of France, on the 
completion of his great work on Theory of Heat, says, "Mathe- 
matics develops step by step, but its progress is steady and cer- 
tain amid the continual fluctuations and mistakes of the human 
mind. Clearness is its attribute, it combines disconnected facts 
and discovers the secret bond that unites them. When air and 



GEOMETRY. 375- 

light and the vibratory phenomena of electricity and magnetism 
seem to elude us, when bodies are removed from us into the 
infinitude of space, when man wishes to behold the drama of 
the heavens that has been enacted centuries ago, when he wants 
to investigate the effects of gravity and heat in the deep, im- 
penetrable interior of our earth, then he calls to his aid the help 
of mathematical analysis. Mathematics renders palpable the 
most intangible things, it binds the most fleeting phenomena, it 
calls down the bodies from the infinitude of the heavens and 
opens up to us the interior of the earth. It seems a power of 
the human mind conferred upon us for the purpose of recom- 
pensing us for the imperfection of our senses and the shortness 
of our lives. Nay, what is still more wonderful, in the study 
of the most diverse phenomena it pursues one and the same 
method, it explains them all in the same language, as if it were 
to bear witness to the unity and simplicity of the plan of the 
universe." 

Mathematics is the very embodiment of truth. No true de- 
votee of mathematics can be dishonest, untruthful, unjust. Be- 
cause working ever with that which is true, how can one de- 
velop in himself that which is exactly opposite. It would be 
as though one who was always doing acts of kindness should 
develop a mean and groveling disposition. Mathematics there- 
fore has ethical value as well as educational value. Its prac- 
tical value is seen about us every day. To do away with every 
one of the many conveniences of this present civilization in 
which some mathematical principle is applied, would be to turn 
the finger of time back over the dial of the ages to the time 
when man dwelt in caves and crouched over the bodies of wild 
beasts. 

The practical applications of mathematics has in all ages re- 
downed to the highest happiness of the human race. It rears 
magnificent temples and edifices, it bridges our streams and 
rivers; it sends the railroad car with the speed of the wind 
across the continent; it builds beautiful ships that sail on every 
sea; it has constructed telegraph and telephone lines and made 
a messenger of something known to mathematics alone that bears 
messages of love and peace around the globe; and by these 
marvellous achievements, it has bound all the nations of the earth 
in one common brotherhood of man. 

IV. AXIOMS. 

7. The self-evident first principles of which mention was 
made in the previous section are called axioms. 

Thus, A can not be both B and non-./? at the same time; A horse is a 
horse; Two times two are four; A body in motion will remain in motion,, 
unless ncted upon by some external force. 

The following are the axioms used in mathematics: 



376 FINKEL'S SOLUTION BOOK. 



GENERAL AXIOMS. 

1. Things equal to the same thing are equal to each other. 

Thus, if A-B and B-C, then A=C. 

2. If equals are added to equals, the sums are equal. 

Thus, if AB and CD, then A+C=B-\-D. 

3. If equals be taken from equals the remainders are equal. 

Thus, if A-B and C=D, then AC=BD. 

4. If equals be added to unequals the sums are unequal in the same order, 
or sense. 

Thus, if A is greater than B and C D, then A-\-C is greater than 
B+D. 

5. If equals be taken from unequals the remainders are unequal in the 
same sense. 

Thus, if A is greater than B and CD, then AC is greater than 
B D. 

6. If unequals be taken from equals the remainders are unequal in the 
opposite sense. 

Thus, if A is greater than B and C is equal to D, then C A is less 
than D B. 

7. If equals be multiplied by equals, the products are equal. 

Thus, \tA=B and C-D, then AC=BD. 

8. If unequals be multiplied by equals, the products are unequal in the 
same sense. 

Thus if A is greater than B and C=D, then AC is greater than BD. 

9. If equals be divided by equals, the quotients are equal. 

Thus, if A=B and C=D, then ---p- 

10. If unequals be divided by equals, the quotients are unequal in the same 
sense. 

A B 

Thus, if A is greater than B and C=D, then is greater than -=^ 

11. If unequals be added to unequals, the greater to the greater and the 
lesser to the lesser, the sums will be unequal in the same sense. 

Thus, if A is greater than B and C greater than D, then A-\-C is 
greater than B-\-D. If m is less than n and p less than q, then 
m-\-p is less than n-\-q. 

12. The whole is greater than any of its parts. 

Thus, if #!, # s > a z ,a are parts of A, then A is grerter than any of 
the a's. 

13. The whole is equal to the sum of all its parts. 

Thus, if lf a 2 , a a , 4 , 5 are the parts of ^4, then A=a 1 -\-a 9 -}-a a -\- 

14. Magnitudes which coincide with one another are equal to one another. 

Thus, if A coincides with B, then A and B are equal. 

15. If of two unequal quantities, the lesser increases continuously and in- 
definitely while the other decreases continuously and indefinitely they 
must become equal once and but once. 

Thus, if, in the figure, the 

line EF moves parallel . n 

to itself, keeping its ex- ~^>^r ** 

tremities in AB and 
MN and the line GH 
moves parallel to itself /j^ 
keeping its extremities 
in MN and CD, then 




, 

the two lines are equal ^' * 

once and only once, 
viz., when both are equal to the line IK. 



GEOMETRY. 37T 

16. If of three quantities the first is greater than the second and the second 
greater than the third, then the first is greater than the third. 

Thus, if A is greater than B and B greater than C, then A is greater 
than C. 

17. Two straight lines can not inclose a \_finite~\ space. 

V. ASSUMPTIONS. 

8. In addition to the definitions of geometrical magnitudes* 
and the above axioms the following Assumptions, or Postu- 
lates, are needed : 

(a.) ASSUMPTIONS OF THE STRAIGHT LINE. 

(i.) One and only one straight line may be passed through 
every two points in space; or, briefly, two. points determine a 
straight line. 

(2.) Two straight lines lying in a plane, determine a point. 

If the two lines are parallel, we still say, for the sake of generality and 
in harmony with conventions adopted in modern geometry, that the two 
lines intersect in a point, the point infinity. By taking this view of two 
parallel lines, many theorems are stated and proved without exceptions to 
either statement or proof. 

(3.) Through any point in space a line may be drawn and 
revolved about this point as a center so as to include any assigned 
point. 

(4.) A straight line-segment, or a sect, may be produced so 
as to have any desired length. 

(5.) A straight line is divided into two parts by any one 
of its points. 

(b.) ASSUMPTIONS OF THE PLANE. 

(i.) Three points not in the same line determine a plane. 

(2.) A straight line through two points in a plane lies wholly 
in the plane. 

(3.) A plcMie may be passed through a straight line and re- 
volved about it so as to include any assigned point in spac&. 

(4.) A portion of a plane may be produced to any desired 
extent. 

(5.) A plane is divided into two parts by any of its straight 
lines. 

(6.) A plane divides space into two parts. 

(c.) ASSUMPTION OF PARALLEL LINES. 

( i.) Through a point without a straight line, only one straight 
line can be drawn parallel to that lin&. 

This assumption is a substitute for Euclid's famous eleventh 
(also called the twelfth) axiom which reads, // a straight line 
meet tzvo straight lines so as to make the iivo interior angles 

*For definitions of geometrical magnitudes, see Mensuration. 



#78 FINKEL'S SOLUTION BOOK. 

on the same side of it taken together less than two right angles, 
these straight lines being continually produced shall at length 
meet on that side on which are the angles zvhich are less than, 
two right angles. 

An axiom must possess the following properties : ( I ) must 
be self-evident, (2) must be incapable of being proved from 
other axioms. That the above so-called axiom does not pos- 
sess the first of these requisites is proved by the fact that there 
is a dispute among mathematicians as to whether it is an axiom 
or not. However, it does satisfy the second criterion as, so 
far, no valid proof of it from other axioms has ever been given. 
Many proofs have, indeed been given, but it requires very little 
thought to see that these proofs are all fallacies of Petitio 
Principii. 

The many attempts to give a rigorous and valid proof of this 
assumption, for such it is, has redounded to the eternal glory of 
geometry in that not only is Euclidean Geometry preserved in 
all its original purity and integrity but other geometries equally 
cogent and consistent have been created. 

The subject is too abstruse for my present purpose and so I 
shall do nothing more than show the point of departure of these 
geometries. 

1. Let AB be a 
given straight line, 
and P the given 
point. 

2. Through P 
draw any number 
of lines. 

3. These lines, 
in relation to the 

given line, divide themselves into two classes, viz., CUTTING and 

NON-CUTTING. 

Now of the class, non-cutting, how many lines are there? On 
the answer to this question "hangs all the law and the prophets." 
A priori, three answers are possible, viz., none, one, many. If 
we say "none," we have Spherical Geometry; if we say 
"one," we have Euclidean Geometry; if we say "more than 
one," we have Pseudo- Spherical Geometry. 

It is true that the answer, "one," is the answer that is usually 
insisted upon as being the only possible answer. But this an- 
swer is based upon experience and is not, therefore, a priori. 

In these geometries, the properties of figures are studied, which 
figures lie in space, or surfaces, possessing the property that 
the product of the principal radii of curvature at every point 
of the surfaces shall be constant. If this product is positive, the 
surface is spherical and the geometry treating of the figures of 
this surface is Spherical Geometry; if this product is 0, the 




GEOMETRY. 379 

surface is a plane, and the geometry treating of the properties 
of figures lying in this surface is the ordinary Euclidean Geom- 
etry ; if this product is negative, the surface is pseudo-spherical 
and the geometry treating of the properties of this space is_ 
Pseudo-Spherical Geometry. 

In the above discussion, it has been assumed tacitly that the 
measure of a distance remains everywhere the same. Professor 
Felix Klein has shown that if this be not the case and if the 
lav/, of measurement of distance be properly chosen, we can 
obtain three systems of plane geometry analogous to the three 
systems mentioned above. These are called respectively Ellip- 
tic, Parabolic, and Hyperbolic Geometries. They mean 
lacking, equaling, and exceeding. Instead of the above terms 
given by Klein, we often meet Riemannian, Euclidean, 
and Isobatschevskian (or Gaussian) from Riemann, Euclid, 
and Lobatschevsky and Gauss, mathematicians who first set 
forth clearly the properties of the space-forms. These geom- 
etries refer to hyper-space of two dimensions and are called 
collectively non- -Euclidean Geometry. 

The notion of hyper-space of two dimensions naturally sug- 
gested the question as to whether there are different kinds of 
hyper-s^>ace of three or more dimensions. Riemann showed that 
there are three kinds of hyper-space of three dimensions having 
properties analogous to the three kinds of hyper-space of two 
dimensions already discussed. These hyper-spaces are differ- 
entiated by the test whether at every point no geodetical surface, 
or one geodetical surface, or a fasciculus of geodetical surfaces 
can be drawn parallel to a given surface, a geodetical surface 
being defined as such that every geodetic line joining any two 
points on it lies wholly on the surface. The student who would 
pursue the subject should read Dr. Halsted's excellent transla- 
tions of Lobatschevsky and Bolyai, the Lectures and Addresses 
of Clifford and Helmholz, Ball's article on Measurement in the 
Encyclopedia Britannica, Professor Schubert's Essay on the 
Fourth Dimension, Russell's Foundations of Geometry, and after- 
wards the monographs of Riemann, Klein, Newcomb, Beltrami, 
and Killing. For a full bibliography of the literature of the 
subject up to the time of its publication, see Bibliography of Non- 
Euclidean Geometry, by Dr. Halsted, American Journal of 
Mathematics. 

(d.) ASSUMPTION OF THE CIRCLE. 

(i.) A circle may be constructed with any point as center, 
and with a radius equal to any given sect. 

(2.) A circle has but one center. 

(3). All radii of the same circle arc equal, and, hence all 
diameters of the same circle are equal. 



380 FINKEL'S SOLUTION BOOK. 

(4.) // an unlimited straight line passes through a point 
within a circle, it must cut the circumference at least twice. 

That it can not cut the circumference more than twice is a 
theorem. 

The region within a circle is defined as that from any point 
of which no tangents can be drawn'to the circle. 

(5.) // one circumference intersects another once, it inter- 
sects it again. 

(e.) ASSUMPTION OF THE SPHERE. 

(i.) A sphere may be constructed with any point as center, 
and with a radius equal to any given sect. 

(2.) A sphere has but one center. 

(3.) All radii of the same sphere are equal, and, hence all 
diameters of the same sphere are equal. 

(4.) // an unlimited straight line passes through a point 
within a sphere, it must cut the surface at least twice. 

(5.) If an unlimited plane or if a spherical surface, intersects 
a spherical surface, it must intersect it in a closed line. 

(/) ASSUMPTION OF MOTION. 

(i.) A figure may be moved from one position in three di- 
mensional space to any other position in the same space without 
altering the size or shape of the figure. 

By this we mean that a figure may be picked up, turned over in any 
way, and moved to any other position in space without changing the size 
or shape of the figure. The proof of many theorems in geometry depends 
upon this assumption. 

(2.) A figure may be moved about in space while one of its 
points remains fixed. 

Such movement is called " rotation about a center," the center being 
the fixed point 

(3.) A figure may be moved about in space while two of its 
points remain fixed. 

Such movement is called " rotation about an axis," the axis being the 
line determined by two fixed points. 

In the higher mathematics and in Physics and other natural 
sciences other assumptions are needed. 

VI. ON LOGIC. 

9. On I/ogic. As a preliminary to the study of geometry 
a short discussion of the Methods of Reasoning will be of value. 

In geometry we are concerned with propositions about 
space relations. Ideas are images of an object formed by the 
mind. Words are the spoken or written signs of ideas. 

10. A judgment is an act of the mind affirming a relation 
between two objects of thought by means of their conceptions. 




U2/V~t^~ 




GEOMETRY. 381 

11. A proposition is a judgment expressed in words. 
For example, take the ideas represented by "all mushroons" 

and "things good to eat," posit these ideas in the mind and dis- 
cern the agreement or disagreement of these two ideas, then 
express the agreement or disagreement in words. It comes out 
thus, 

"All mushroons are things good to eat." 

Our senses are the instruments by which the qualities of a 
mushroon are made known to us. Having found this mushroon 
good to eat, and this one, and this one, and so on, together with 
the experience of the race, we arrive at the conclusion, by in- 
ductive inference, that "all mushroons are good to eat." It must 
be borne in mind that by induction we gain no certain knowledge. 
If the observation of a number of cases shows that alloys of 
metals fuse at lower temperatures than their constituent metals r 
we may with more or less probability draw the general inference 
that 

All alloys melt at a lower temperature than their constituent 
metals. 

But this can never rise to the rank of an absolutely certain 
law until all possible cases have been examined. Not one of 
the inductive -truths which men have established, or think they 
have established, is really safe from exception or reversal. 
Lavoisier, when laying the foundations of chemistry, met with 
so many instances tending to show the existence of oxygen in 
all acids that he adopted the general conclusion that all acids 
contain oxygen, yet subsequent experience has shown this to 
be false. Like remarks may be made concerning all other in- 
ductive inferences, the method never leading to absolute certainty. 

12. The Powers of the Mind engaged in knowledge are the 
following three, viz., 

1 i ) The Power of Discrimination, 

(2) The Power of Detecting Identity, and 

(3) The Power of Retention. 

13. The Laws of Thought are the following three, viz., 

1 i ) The Law of Identity ; as, That which is, is. 

(2) The Law of Contradiction ; as, A thing cannot both 
be and not be at the same time. 

(3) The Law of Duality; as, A thing must either be or 
not be. 

To these some logicians add a fourth called the "Law of Suf- 
ficient reason ;" Every effect has a cause. 

14. When we join terms together we make propositions; 
when we join propositions together we make an argument, or 
piece of reasoning. 



382 FINKEL'S SOLUTION BOOK. 

15. Terms. A concrete term has two meanings, viz., (i) 
things to which the term applies, and (2) the qualities of those 
things in consequence of which the term is applied. The num- 
ber of different things to which a term is applied is called its 
extension, while the number of qualities implied is called its 
intension. 

For example, "table" has a larger "extension" than "round 
table" for the former term applies to a larger number of objects ; 
the latter has the greater "intension" for it includes all the quali- 
ties that the term "table" does and the additional quality "round." 

The word "term" comes from the Latin terminus, meaning 
end and is so called because it forms one end of a proposition. 

1 6. Propositions. Every proposition is composed of a 
subject, ( Lat., sub, under, and jectum, laid), a copula,, and a 
predicate (Lat. praedicare, to assert). 

In the proposition, "All mushroons are things good to eat," 
"all mushroons" is the subject, "are" is the copula, and "things 
good to eat" is the predicate. 

Of the kinds of propositions we have 

(i) Categorical; As A is B. A is not B; (2) Con- 
ditional; as, If a triangle is equiangular, it is equilateral. 

Conditional Propositions are divided into two classes, 
viz., Hypothetical and Disjunctive. The following is a 
disjunctive proposition: 

A is either B or C. 
Of the Categorical Propositions we have, 

A. The Universal Affirmative; as, All horses are 

animals. 
E. The Particular Affirmative; as, Some animals are 

horses. 

I. The Universal Negative; as, No horses are cows. 
O. The Particular Negative; as, Some animals are not 

horses. 

Every proposition which expresses accurately a thought, can 
be reduced to one of the above forms, though the reduction in 
many cases is not apparent. For example, 

Parallel lines never meet, reduces to 
Parallel lines are lines which never meet. 

The hypothetical proposition, "If gunpowder be damp, it will 
not explode" reduces to, "Damp gunpowder will not explode." 

When we make a statement about all the objects which can 
be included under a term, we use the term UNIVERSALLY, as 
logicians say, that is to say, THE TERM is DISTRIBUTED. In the 
proposition, "all men are mortal," the term "men" is distributed, 
because the little word "all" indicates that the statement applies 



GEOMETRY. 383 

to any and every man. But THE PREDICATE "mortal" is ONLY 

TAKEN PARTICULARLY AND IS NOT DISTRIBUTED. 

Therefore, we see that a UNIVERSAL AFFIRMATIVE DISTRIB- 
UTES ITS SUBJECT BUT NOT ITS PREDICATE. 

As a universal negative proposition take, "No sea-weed is a 
flowering plant." The subject "sea-weed" is distributed. If 
there could be found a single flowering plant which is a sea- 
weed, then the proposition would not be true. Hence the predi- 
cate is also distributed. 

Hence, THE UNIVERSAL NEGATIVE PROPOSITION DISTRIBUTES ITS 

SUBJECT AND ITS PREDICATE. 

No difficulty is experienced in seeing that the particular affirma- 
tive distributes neither its subject nor its predicate, and that the 

PARTICULAR NEGATIVE DISTRIBUTES ITS PREDICATE BUT NOT ITS 
SUBJECT. 

In the absence of any knowledge to the contrary, the word 
"some," in the particular affirmative and particular negative, must 
be taken to mean "SOME AND IT MAY BE ALL/' 

17. Tlie I/a w of Converse. Two propositions are the 
converse of each other when the subject of one is the predicate 
of the other. Thus, 

"Equilateral triangles are equiangular, "(direct). 
Equiangular triangles are equilateral, (converse). 

It does not follow that because a proposition is true its con- 
verse will also be true. Thus, "All regular polygons are equi- 
lateral (direct) ; all equilateral (polygons) are regular, (con- 
verse). This last is not true. The converse of all definitions 
are true. 

Whenever three theorems have the following relations, their 
converses are true : 

1. If it is known that when A > B, then x *> y, and 

2. If it is known that when A = B, then x y, and 

3. If it is known that when A < /?, then x < y, 
then the converse of each of these is true. 

For 

l x . If x >y, then A cannot equal B and A cannot be less 
than B without violating 2 or 3; .'. A > B. (Converse of 1.) 

2 18 If x y, then A cannot be greater than B and A cannot 
be less than B without violating 1 or 3; .*. A=B. (Converse 
of 2.) 

3 t . If x < y, then A cannot be greater than B and A cannot 
be equal to B without violating 1 or 2; .'. A < B. (Converse 
of 3.) 

1 8. The opposite of a proposition is formed by stating the 
negative of its hypothesis and conclusions. Thus, 

If A B, then C = D (Direct.) 

If A is not equal B, then C is not equal D. (Opposite.) 



384 FINKEL'S SOLUTION BOOK. 

19. // the direct proposition and its converse are true, the 
opposite proposition is true; and if a direct proposition and its 
opposite are true, the converse proposition is true. Thus, 

1. It A=B, C = D. (Direct.) 

If C = D, A = B. (Converse.) 

If A is not equal to B,-C is not equal to D (Opposite.) 

2. If A = B, C = D. (Direct.) 

If A is not equal to B, C is not equal to D. (Opposite.) 
Then, if C = D, A = B. (Converse.) 

20. Methods of Reasoning. There are two methods of 
reasoning, viz., the Inductive and the Deductive. 

The Inductive Methodis used in reaching a general truth 
or principle by an examination and comparison of particular 
facts. Thus, This apple is equal to the sum of all its parts, this 
piece of crayon is equal to the sum of all its parts, this orange 
is equal to the sum of all its parts, and so with peaches, pears,, 
balls, pebbles, slates, knives, and chairs. 

Therefore, the whole of any object is equal to the sum of all 
its parts, or the whole is equal to the sum of all its parts. This 
is inductive reasoning. 

The Deductive Method is used in reaching a particular 
truth or principle from general truths or principles. Thus. 

All animals suffer pain. 
Flies are animals. 
Therefore, flies suffer pain. 

21. Tie Syllogism. When we compare propositions we 
reason. Deriving a third proposition from two given proposi- 
tions is called syllogistic reasoning, or Deductive Rea- 
soning. Thus, 

1. All English silver coins are coined at Tower 

Hill. 

2. All sixpences are coined at Tower Hill. 
Therefore, All sixpences are English silver coins. 

The last proposition is called the conclusion, the other two 
propositions are called premises, and the three together the 
syllogism. 

Again, 

All electors pay rates. A. 

No paupers pay rates. E. 

Therefore, no paupers are electors. E. 

From the examples given, we see that there are only three 
terms or classes of things reasoned about ; in the first example 
the three terms are "All English silver coins/' "Tower Hill," and 
" all sixpences." Of these, the class, " English silver coins," 
does not occur in the conclusion. It is used to enable us to 
compare together the other two classes of things. It is called 




RENE DESCARTES. 



GEOMETRY. 385 

the middle term. (Things) "coined at Tower Hill," is called 
the major term for the reason that it has the larger exten- 
sion, and "sixpences," the subject of the conclusion, is called the 
minor term of the syllogism, for the reason that it has a lesser 
extension than the subject of the conclusion. 

The premise in which the "major term" is found is called the 
major premise, and the one in which the minor term is found 
is called the minor premise. 

Hence, the middle term is always the term not found 
in the conclusion ; the major term is the predicate of 
the conclusion ; and the minor term is the subject of 
the conclusion. 

Suppose that the two premises and the conclusion of the last 
syllogism be varied in every possible way from affirmative to 
negative, from universal to particular and vice versa. 
. Each proposition can be converted into four different propo- 
sitions and each one of these four may be compounded with any 
one of the other two. Hence the number of changes (called 
moods) is 4 X 4 X 4 64. These moods may be still further 
varied, if instead of the middle term being the subject of the 
first and the predicate of the second, this order may be reversed, 
or if the middle term the subject of both, or the predicate of 
both. In this way we see that for each of the sixty-four moods 
we get four syllogisms called figures. 

Of the sixty-four moods, there are altogether nineteen 
moods of the syllogism that are admissible. 

22. Rules of the Syllogism. To find out whether an 
argument is valid or not, we must examine it carefully to ascer- 
tain whether it agrees with certain rules discovered by Aristotle. 
Modern logicians have to some extent broken away from these 
rules. Without going into the matter in detail we state these 
rules. 

I. Every syllogism has three terms and only three. 
These terms are called the the major term, the minor term, and 
the middle term. 

II. Every syllogism contains three and only three 
propositions. 

III. The middle term must be distributed once at 
least in the premises and must not be ambiguous. 

Some animals are flesh-eating. 
Some animals have two stomachs. 
No conclusion can be drawn. 



But if we say, 

Some animals are flesh-eating, 
All animals consume oxygen, we 
can say 

Therefore, some animals consuming oxygen are flesh-eating. 




FINKEL'S SOLUTION BOOK. 



IV. If both premises are negative no conclusion 
can be drawn. 

For, from the statements that two things disagree with a third, 
no proof of agreement or disagreement can 
be established. Thus the following is incon- 
clusive, 

No Americans are slaves. 

No Turks are Americans. 

V. If both premises are particular 
no conclusion can be drawn. 

Thus the following are inconclusive: 

Some Americans are ignorant. 
Some Europeans are ignorant. 

Some laws are unjust. 
Some men are unjust. 

VI. No term must be distributed in the conclusion 
which was not distributed in the premises. 

From 

Some animals eat flesh. 

All animals consume oxygen. 

We must conclude that some things that 
consume oxygen eat flesh. 






VII. If one premise be negative the conclusion 
must be negative. 

Thus from 

All negroes are dark. 
No American is dark. 

We draw the conclusion 

No American is a negro. 

VIII. If either premise is particu- 
lar the conclusion must be particular. 

Thus, 

All negroes are black. 
Some horses are black. 

Therefore, some horses are not negroes. 




23. logical Fallacies. Logical Fallacies result from our 
neglect to observe the rules of logic. They occur in the mere 
form of the statement, that is, in dictione, as it is known in logic. 



GEOMETRY. 387 

There are four purely logical fallacies, viz., 

1. Fallacy of four terms (Quaternio Terminorum), 

Violation of Rule I. 

2. Fallacy of undistributed middle, Violation of Rule 

III. 

3. Fallacy of illicit process, of the major or minor term. 

Violation of Rule VI. 

4. Fallacy of negative premises. Violation of Rule 

IV. 
There are six semi-logical fallacies, viz., 

1. Fallacy of Equivocation. 

2. Fallacy of Amphibology. 

3. Fallacy of Composition. 

4. Fallacy of Division. 

5. Fallacy of Accent. 

6. Fallacy of Figure of Speech. 

In addition to these logical fallicies there are seven Material 
Fallacies (extra dictionem)tha.t is, fallacy in the matter of 
thought, viz., 

1. Fallacy of Accident. 

2. The Converse Fallacy of Accident. 

3. The Irrelevant Conclusion. 

4. The Petitio Principii. 

5. The Fallacy of the Consequent or Non-sequitur. 

6. The False Cause. 

7. The Fallacy of Many Questions. 
We will illustrate some of these fallacies. 

Light is contrary to darkness. 
Feathers are light. 
.-. Feathers are contrary to darkness. 

The middle term, "light," has two different meanings in the 
premises. We have, therefore, four terms instead of three, which 
violates Rule I. When the middle term is ambiguous, the fal- 
lacy is known as the ambiguous middle. 

Every country under a tyranny is distressed. 
This country is distressed. 

.-. This country is under a Tyranny. Fallacy of Undis- 
tributed Middle. 

All moral beings are accountable. 
No brute is a moral being. 

.-. No brute is accountable. Fallacy of the Illicit Process 
of the Major Term. 

Some men are not just. 
No angel is a man. 
.-. Some angels are not just. Fallacy of Negative Premises. 



388 FINKEL'S SOLUTION BOOK. 



EXAMPLES. 

Seven is one number. 
Two and five are seven. 
.-. Two and five are one number. Fallacy of Division. 

Three and four are two numbers. 
Seven is three and four. 
.-. Seven is two numbers. Fallacy of Composition. 

The duke yet lives that Henry shall depose. Fallacy of 
Am philology. 

The conclusion depending upon the interpretation of the mean- 
ing of this proposition is doubtful. 

A hero is a lion. 
A lion is a quadruped. 
.-.A hero is a quadruped. Fallacy of Figure of Speech. 

Thieves are dishonest ; 
But thieves are men; 
.-. All men are dishonest. Fallacy of Accident. 

24. How to Prepare a L,esson in Geometry. In be- 
ginning the study of geometry, great care should be taken to grasp 
a correct notion of the definitions and illustrations. The defi- 
nitions, axioms, and assumptions are the foundation on which 
rests the magnificent structure of geometry. The definitions 
should be committed to memory, only committing them, however, 
as they occur in the prosecution of the study. Make haste slowly 
at first ; one proposition per lesson for the first three lessons 
is quite sufficient; and two propositions may be taken at a les- 
son for the next seven or eight lessons. After this, if the work 
is thoroughly in hand three propositions together with several 
originals should constitute a lesson. 

In the preparation of the lesson, the student should carefully 
read the proposition so as to get its full meaning. After the 
meaning of the proposition is understood, carefully follow the 
demonstration in the book, never leaving a statement made in 
the demonstration until it is thoroughly understood. At first, 
it may be necessary to repeat this two or three times, perhaps 
oftener. After the given demonstration is thoroughly under- 
stood, close the book, draw a figure on paper or a slate, and 
write out a demonstration of your own. Compare your demon- 
stration with the one in the book, and make such corrections as 
are necessary. 

By carefully observing this method, it will be a comparatively 
short time until one reading of the lesson will generally suffice 
for the necessary preparation. The theorem should always be 




LEONHARD EULER. 



GEOMETRY. 



389 



committed to memory, the demonstration never. It is not a bad 
practice to commit the proposition exactly as it is stated in the 
book, for, as a general thing the author has put much time on 
the statement of each proposition endeavoring to reduce it to 
its simplest and most elegant form, and upon this work, the 
student, as a rule, can not improve. 

In conducting the recitations, no books should be allowed to 
be consulted. The propositions should be assigned by stating 
them in part or in full to the students called upon to recite. The 
students so called upon, should go to the board and draw as 
neat and accurate figure as possible, accurate figures often sug- 
gesting truths not revealed by carelessly constructed figures. It 
is generally best not to require any part of the demonstration to 
be written out, unless, indeed, it includes long and complicated 
algebraic equations. In reciting, if it is convenient, the student 
should step to the board and, using a pointer in referring to the 
various parts of his figure, observe the following order in the 
discussion of the theorem : 

I. Statement of the Theorem. Here give an accu- 
rate statement of the theorem to be demonstrated. 
II. Given. Here state, with reference to the figure con- 
structed whatever is given by the theorem. 

III. To Prove. Here state the exact conclusion to be de- 

rived from what is given. 

IV. Proof. Here set forth, in logical order the statements 

to prove the conclusion just asserted. 

The validity, limitations, and general application of the the- 
orem may then be discussed by the class. 

Corollaries coming under the various theorems in the lesson 
may be assigned to students other than those demonstrating the 
theorems. The proof of a corollary is usually simple, but its 
proof should be given with the same care and accuracy. 

We will now illustrate what we have said by a few proposi- 
tions. The student should have one of the following excellent 
texts : 

Halsted's Elements of Geometry.. 

Beman and Smith's Plane and Solid Geometry. 

Phillips and Fisher's Elements of Geometry. 

Wentworth's Plane and Solid Geometry. 



390 



FINKEL'S SOLUTION BOOK. 



PLANE GEOMETRY. 



BOOK I. 



ANGLES AND STRAIGHT LINES. 



I. Theorem. 



PROPOSITION I. 

All straight angles are equal. 



C 

I 



.B 



I. 



II. 
III. 



E 

I. 



.F 



II. Given any two straight angles ACB and DEF. 
III. To prove /_ ACB = / DEF. 

f 1. Apply / ACB to the / DEF, so that the 

vertex C shall fall on the vertex E. 
(First assumption of motion.) 

2. Then revolve CB so that it contains the 

point F. 
(Third assumption of the straight line.) 

3. Then CA will coincide with ED. 

(First assumption of a straight line and 

Law of Identity?) 
14. .'. tACB = /_DEF. Axiom 10. 

Corollary 1. All right angles are equal. 
A A 



IV. Proof. 



C 
Given 



B 



C 



any two right angles ACB and A C' B f . 
To prove /_ ACB = / ACB'. 



1. All straight angles are equal. Prop. I. 

IV Proof. \^' ^- ACB . and / ^C'ff are each the half of a 

straight angle. By definition. 
3. .-. l_ACB = tAC'B'. Axiom?. 



GEOMETRY. 



391 



I. Cor. 2. The angular units, degree, minute, and sec- 
ond have constant values. 

II. Given a degree angle. 
III. To prove that it is a constant magnitude. 

1. A constant magnitude is a magnitude 

By 



IV. Proof. 



whose value is always the same, 
def. 

A straight angle is a magnitude whose 
value is always the same. By Prop. I. 

1. .'.A straight angle is a constant mag- 

nitude. 

2. A degree angle is one hundred eight- 

ieth part of a straight angle. By 
def. 

3. .'.A degree angle is a constant mag- 

nitude. By Aristotle's Dictum, 
Whatever may be predicated of a 
whole may be predicated of a part. 

In like manner, we can prove that minute-angles and second- 
angles are constants. 

I. Cor. 3. Complements of equal angles are equal. 




IV. Proof. 



II. Given the two equal angles CBD and CB'tf and 
their complements ABC and A B' C ', respectively. 

III. To prove that / ABC = / A'B'C. 

/ ABC = the difference between a rt. / 

and / CBD. By def. of comp. 
/_ A'B'C' = the difference between a rt. / 
" and / C'B'jy. By def. of comp. 
But / CBD = / C'B'Df. By hypothesis. 
/. / ABC = / A'B'C. By Axiom 1. 

I. Cor. 4. Supplements of equal angles are equal. 
(Proof same as above.) 

I. Cor. 5. At a given point in a given line, one perpen- 
dicular, and only one, can be erected in the same 
plane. 



392 



FINKEL'S SOLUTION BOOK. 




II. Given 



,-r '/> 

CD perpendicular to AB at P. 



IV. Proofs 



III. To prove that no other perpendicular can be drawn to 
AB at P in the same plane. 

fl. Suppose that another perpendicular EF 
could be drawn. 

2. Then / BPE would be a rt. /. By def. of 

perpendicular. 

(If two lines meet and form a rt. L, each is said to be per- 
pendicular to the other.) 

3. But / BPC is a rt. angle. 

(Since CD is perpendicular to AB.) 

4. /; 2 BPE would equal / BPC. Prop. I., 

Cor. 1. 

(All right angles are equal.) 

5. But this is impossible. By Axiom 8. 

(The whole is greater than any of its parts.) 

6. .'. The supposition of step 1 is absurd, 

and a second perpendicular is impossi- 
ble. Q. E. D. 

Remark. In this demonstration, we have used what is called 
the Indirect Method, or reductio ad absurdum which means 
a reduction to an absurdity, as distinguished from the Direct 
Method used in the other proofs. Jevons in his Principles of 
Science, Vol. I, p. 96, says, "Some philosophers, especially those 
of France, have held that the Indirect Method of Proof has a 
certain inferiority to a direct method, which should prevent our 
using it." He goes on to show that the method is not inferior 
and holds the belief that nearly half our logical conclusions rest 
upon its employment. 

In the above case, by the Law of Duality, a second perpen- 
dicular can or can -not be drawn. It was shown that by sup- 
posing that a second one could be drawn led us to an absurdity. 
Hence, a second can not be drawn. This method of proof is 
often used in geometry. 

PROPOSITION II. 

I. Theorem. If two adjacent angles have their exterior 
sides in a straight line, these angles are supplements 
of each other. 







SOPHUS UK. 




393 




B 

\j 

II. Given the exterior sides OA and OB of the adjacent 
angles AOD and BOD respectively and the straight 
AB in which these two sides lie. 
III. To prove / AOD = / DOB. 

AOB is a straight line. By hypothesis. 
.'. / AOB is a st. /. By def. of a st. /. 
Kut/_AOD + /_DOB = /_AOB. By Ax. 9. 
.*. /'s AOD and Z>&# are supplementary. 
By def. of supl. angles. 

Cor. i. The sum of all the angles about a point in a plane 
is equal to two straight angles. 

Cor. 2. The sum of all the angles about a point on the same 
side of a straight line passing through a point, is equal 
to a straight angle: 

PROPOSITION III. 

I. Theorem. CONVERSELY: // two adjacent angles are 
supplements of each other, their ez. terior angles lie in 
the same straight line. 



B 




II. Given that the sum of the adjacent angles AOD 
and DOB are supplements of each other, that is, 
equal to a straight angle. 
III. To prove AO and OB in the same straight line. 

1. Assume OF in the same straight line with 

OA. 

2. Then /_AOD + /_DOF\$z straight angle. 

By Prop. II. 

3. But / AOD + / DOB is a straight angle. 

By hypothesis. 

4. /. /_AOD + /_DOB = /_AOD + /_DOF. 



IV. Proof. < 



By Ax. I. 

5. AOD= /_AOD. By Law of Identity. 

6. Subtracting step 5 from step 4, / DOB - 

/ DOF. By Ax. 3. 



394 



FINKEL'S SOLUTION BOOK. 



7. .'. OB and OF coincide. By converse 

Ax. 10. 

8. .'. AO and OB are in the same straight 

line. Q. E. D. 
SCHOLIUM. Since Propositions II. and III. are true, their 

opposites are true, viz., 

// the exterior sides of two adjacent angles are not in a straight 
line, these angles are not supplements of each other. 

If two adjacent angles are not supplements of each other, their 
exterior sides are not in the same straight line. 



I. 



PROPOSITION IV. 

Theorem. If one straight line intersects another 
straight line, the vertical angles are equal. 




IV. Proof. 4 



II. Given the two lines AB and DE intersecting in 0. 
III. To prove /_ AOE = /_ DOB. 

1. / AOE + /_ AOD equals a st. /.By 
" Prop. I. 
/ AOD + / DOB equals a st. /. By 

Prop. I. 
. . / AOE + / AOD = / AOD + / Z?O. 

By Ax. 1. 
Take away from each of these equals the 

common / AOD. 

Then / AOE = /_ DOB. By Ax. 3. 
In like manner we may prove / AOD = 

/_ EOB. Q. E. D. 
I. Cor. If one of the four angles formed by the intersec- 
tion of two straight lines is a right angle, the other three 
angles are right angles. 

PROPOSITION V. 

I. Theorem. From a point without a straight line one 
perpendicular, and only one, can be drawn to this line. 



2. 



GEOMETRY. 



395 



IV. Proof. 4 



II. Given the point, P, and the straight line, AB. 

III. To prove that one perpendicular can be drawn from 
P to AB, and only one. 

1. Turn the part of the plane above AB about 

AB as an axis until it falls upon the part 
below AB and denote the position of P 
by P r . By Assumption 3 of the Plane. 

2. Turn the revolved plane about AB to its 

original position. By Assumption 3 of 
the Plane. 

3. Draw the straight line PP', cutting AB in 

C. By Assumptions 1 and 2 of the 
Straight Line. 

4. Take any other point D in AB, and draw 

PD and P'D. 

5. Since PCP' is a straight line, PDP' is not 

a straight line. 

(Between two points only one straight line can be drawn.) 

6. Turn the figure PCD about AB until P 

falls on P'. By Assumption 3 of the 
Plane. 

7. Then CP will coincide with CP' and DP 

with DP'. 

8. /. / PCD = /_P'CD, and / PDC = / 

P'DC. Ax. 15. 

9. .*. / PCD, the half of a St. / PCP' is a 

right /; and / PDC, the half of / PDP', 
is not a right angle. 

10. .*. PC is perpendicular to AB, and />/? is 

not perpendicular to AB. By def. of 
Perpendicular. 

11. .'. one perpendicular, and only one, can be 

drawn from P to AB. Q. E. D. 

PARALLEL LINES. 

Definition. Parallel lines are lines lying in the same plane 
and never meeting however far produced. 

On this definition and the assumption of parallel lines rests 
the whole theory of parallel lines in Euclidean geometry. By 
convention, we say that parallel lines meet at infinity. Why this 
convention is adopted will become apparent in studying Higher 
Modern Geometry. 

PROPOSITION VI. 

I. Theorem. Two straight lines in the same plane per- 
pendicular to the same straight line are parallel. 



396 



FINKEL'S SOLUTION BOOK. 



B 
D 



II. Given the two straight lines AB and CD each per- 
pendicular to the straight line AC. 

III. To prove AB and CD parallel. 

1. AB and CD, lying in the same plane, must 

either meet or not meet. By Law of 
Duality. 

2. If they meet, we shall have two lines from 

the same point perpendicular to the 
same line. By hypothesis. 

IV. PrOOf. { (The lines AS and CD being perpendicular to AC.) 

But this is impossible. By Prop. V. 

(From a given point without a straight line, one perpen- 
dicular, and only one, can be drawn to a straight line.) 

. .'. AB and CD cannot meet, however far 

produced. 
. .'. AB and C>are parallel. By definition 

of Parallel Lines. 



PROPOSITION VII. 

I. Theorem. If a straight line is perpendicular to one of 
two parallel lines, it is perpendicular to the other. 
H 



M . 



c 



D. 



N 



K 



II. 



Given the parallel lines AB and CD and the line 

//^perpendicular to AB. 

III. To prove that HK is perpendicular to CD. 

1. Suppose MN drawn through ^perpendic- 

ular to HK. 

2. Then MNis parallel to AB. By Prop. VI. 

(Two lines in the same plane perpendicular to the same 
line are parallel.) 

^' But ^ D is P arallel to AB. By hypothesis. 

4. . ' . MN coincides with CD. By assump- 

tion 1 of parallel lines. 

(Through a point without a straight line only one straight 
line can be drawn parallel to that line.) 

5. CD is perpendicular to HK\ that is, 
16. HK'vs perpendicular to CD. Q. E. D. 



IV Proof 




SIMON NEWCOMB, PH. D., LL. D. 



GEOMETRY. 397 



TRANSVERSALS. 

Definition. A straight line intersecting two or more straight 
lines is called a transversal. 




In the figure EF is a transversal of the two non-parallel lines 
AB and CD. 

The angles AHI, BHI, CIH, and DIH are called interior 
angles, and the angles AHE, EHB, GIF, and FID are called 
exterior angles. 

The angles ^/// and HID, or 5J77 and H/C are called alter- 
nate-interior angles. 

The angles AHE and >/F, or BHE and C/F are called alter- 
nate-exterior angles. 

The angles AHE and C7tf, ^/// and CIF f EHB and #/D, or 
BHI and Z7/F are called exterior-interior angles. 

PROPOSITION VIII. 

I. Theorem. If two parallel lines are cut by a third 
straight line, the alternate-interior angles are equal; 
and conversely. 



I. Find the value of an angle (1) if it is double its complement; (2) if 
it is one-fourth of its complement. 

II. Given (1) that / A is double its complement. 
III. To find the value of / A. 

1. rt. / / A complement of / A. By def. 

of compl. 

2. A = 2(rt. / / A). By hypothesis. 

3. .*. / A = 2 rt. /'s 2 / A. By Distribu- 

tive Law of Multiplication. 

4. Adding 2 / A to these two equals, we have 

3 / A = 2 rt. /'s. By Ax. 2. 

5. .-. A = %rt. /. By Ax. 7. Q. E. F. 

Let the student give the solution of (2). 

2. Find the value of an angle (1) if it is three times its supplement; (2) if it is 
one-third of its supplement. 

3. How many degrees in the angle formed by the hands of a clock at 2 o'clock? 
3 o'clock? 4 o'clock? 9 o'clock? 



IV So/nf ion <! 
m> { 



398 FINKEL'S SOLUTION BOOK. 

PROPOSITION IX. 

I. Theorem. If two parallel lines are cut by a third 
straight Ime, the exterior angles are equal, and con- 
versely. 

Let the student give the demonstration and state and prove 
the corollaries, if any, coming under the theorem. 





GEORGE BRUCE HAIySTED. 



GEOMETRY. 



GEOMETRY. 



1. Geometry is the science that treats of position ana 
extension. 



2. Pure Geometry 



1. Plane. 

2. Solid. 



line drawn from 




Problem. -To bisect a given triangle by 
a random point in one of its sides. 

Demonstration. Let ABC be the given triangle. D a random 
point in the side BC, and E the middle 
point of BC. Join A and D, A and E. 
Draw EF parallel to AD. Draw DF. 
Then DF bisects the triangle ABC. 
For the triangle ABE is equivalent to 
the triangle AEC (?). The triangle 
AFD is equivalent to the triangle FIG. 7. 

ADE (?). Hence, ABDF is equivalent to ABE (?)and, 
therefore, DF bisects the triangle^ BC. Q. E. D. 

Proposition. The square described upon the hypotenuse of 
a right triangle is equal to theszim of the squares of the other two 
sides. 

I. Demonstration. Let CFD be any right triangle, right an- 
gled at F and let A C, CP, and DM 
be the squares described upon its 
sides. Then the square A C is equal 
to the sum of the squares CP and 
DM. Through F, draw ^/^per- 
pendicular to AB and produce it to 
meet OP produced, in G\ also pro- 
duce BC to meet OP in /and AD 
to meet OP produced, in R. Draw 
GH parallel to PD, and BT par- 
allel to CF. Draw AE. Now the 
triangles COI and D FC are 
equal ( ? ) . Hence , C/= CD= CB, 
and therefore the square CJ D =the 
parallelogram CG (?)=the paral- 
lelogram BE (?)=the rectangle FIG. 2. 
BK (?). In like manner, the square DM can be proved equal 
to the rectangle AK. Hence, the square A C=the square CP-\- 
the square DM. Q. E. D. 




400 



FINKEVS SOLUTION BOOK. 



II. Demonstration. Let EDC\)Q any right triangle, right 
angled at D. On the sides DE and 
DC construct the squares EDHG and 
DCBM respectively. Produce GE 
and BC until they meet in F, form- 
ing the square F B A G. On EC, 
the hypotenuse, construct the square 
ECKI. Then the square ECKI is 
equal to the sum of the squares EDHG 
and D CBM. For, the square GFBA 
is equal to GEDH+DCBM+ 
2 EDCF (=ECF). The square 
GFBA is also equal to the square 
Hence, ECKI+ 
GEDH-\-DCBM+ECF (?). 
= GEDH+D CBM. Q. E. D. 




FIG. .3 

Whence, 



ECKI 



Proposition. In any triangle, each angle formed by join- 
ing the feet of the perpendiculars is bisected by the perpendicu- 
lar from the opposite vertex. 



Demonstration. Let ABC be any triangle and AD, BE, and 
CF the three perpendiculars. Join D and E, D and F, and E 
and F. 

In the right triangles AEB and AFC, the angle BA C is 
common to both. Therefore, they are similar. Hence, AB\A C 
=AE\AF. Now the triangles BA C and FAE have the angle 
FAE common and the including sides proportional. Therefore, 
they are similar, and the angle AFE=\\\e angle ACB. In a 
similar manner we may prove that the angle DFJ3=\he angle 
A CB ; the angle A FE=the angle DFB. 
From this it follows that the angle C FA 
the angle EFA=the angle CFB 
the angle DFB. Hence, angle EFC=snc\- 
gle CFD and the angle EFD is bisected 
by the perpendicular C F. In a similar 
manner, it can be proved that A D bisects 
the angle FDE and EB bisects the angle 
FED. q. E. D. FIG. 4. 

Problem. From a given point in an arc less than a semi- 
circumference, draw a chord of the circle which will be bisected 
by the chord of the given arc. 

Demonstration. Let ABDC be the given circle, AB the 
given arc, AB the chord of the arc, and P any point of the arc 





PROFESvSOR FELIX KLEIN. 



GEOMETRY. 



401 




FIG. 5. 



A PC. Draw the diameter POC and on the radius PO as a di- 

ameter describe the o\\c\?,PEO. Then 

through the points^, and G, of inter- 

section draw the chords PD and PF 

respectively, and they will be bisected 

at the points E and G. For draw 

DC and OE. Then the triangles 

PJZOand PD Care right triangles(?) 

and are also similar (?). Since PEO 

and PD Care similar, the line OE 

is parallel to DC, and since O is the 

middle point of P C,E is the middle 

point of P-D( ?). In like manner, G 

is the middle point of PF. Q.E. P 

Discussion. There are, in general, two solutions. When arc 
AB is diminished until B coincides with A, there is no solution. 
When AB is a semi-circumference, there is one solution and the 
chord is the diameter POC. 

Proposition. If two equal straight lines intersect each 
other anywhere at right angles, the quadrilateral formed by join- 
ing their extremities is equivalent to half the square on either 
straight line. 

Demonstration. Let AB and CD be two equal straight lines 
intersecting each other at right angles 
at E. Join their extremities, form- 
ing the quadrilateral A CBD. Then 
ACBD is equivalent to half the 
square of AB or CD. For the area 
of the triangle A CB equals \(AB 
X CD) and the area of the triangle 
ADB equals ^(ABxED). Hence, 
the area of A 



FIG. 6. 



A PROBLEM IN MODERN GEOMETRY. 

An equilateral hyperbola passes through the middle points 
D, E, and F of the sides BC, AC, and AB of the triangle 
ABC, and cutting those sides in order in a, ft, and y. Show 
that the lines Act, Bft, and Cy intersect in a point the locus of 
which is the circumscribing circle of the triangle ABC. 

Solution. The equation to any conic is ua 2 -\-vfi 2 -\-ivy ' 2 -f- 
Zu'fiy+Zv'ay+Zw'aft^O .... (1). D is (0, %a sin C, \a sin^) ; 
E (i^sinC, 0, UsinA) ; F, (-Jcsin B, cs\nA,0). These 



by 



But CD equals AB, 
hypothesis. Hence, ACBD 
Q. E. D. 




402 FINKEL'S SOLUTION BOOK. 

points being on (1), we should have c z v-{-b 2 TU-\-2bcu'=0 ... (2), 
ctu+a^w+Zacv'^-O ..... (3), t>*t4+a 2 v + 2al>iv' = ....(4). 

Whence u=~(au' bv / CTV') .... (5), v= (bv' civ' au') . . 
0c v ac^ 

. . . (6), w= (cw x au' bv') .... (7). Substituting in the con- 



dition w-fv+w 2z/ cosA2v' cos.# 2w'cosC=0 ....... (8> 

that (1) is an equilateral hyperbola, 



abc 

1v' cos 2<w'cosC0 ..... (9). Clearing of frac- 
tions and noticing that 2abc cosA=a(t> 2 -{-c 2 a 2 ) ..... (10), 
^abcco^=b(a z +c^ 3 2 ) ____ (11), 2a6ccos C=c(a*+b 2 ^} 
..... (12), and reducing, / cos^4-i; / cos^-)-w / cosC=0 ---- (13). 
Substituting (5), (6), and (7) in (1) an clearing of fractions, 



+y 2 +2u'a&c/3y+2v'a&cay-\-'2'w'aZ>ccx/3=Q . . . (14). Where this- 



cuts BC, a=0, and (14) gives &*(&/ cw' au') 



= C Z (CTV' au' bv') . . . (15), whence for the point a; a l 

~ ' By s y mmetr y for the point. 



The 



is found to be ( cw x +^v x aw')/3 (cw x bv' au')y=0 . 
. . . (18) ; to Bft, a(c-w'+au'bv')ac(civ'au'bv')y=0. 
---- (19) ; and to Cy, b(au'+bv'c>w'}p 
a(au' bv' civ')a=0 .... (20) r any two of which meet in 



A ,_ac ( cw'bv'au' ) (c-w'+au'bv' ) 

r* ~^r 

, ab(cw'+bv'au')(c'w'+au'bv') 

~~DT 
The circumscribing circle is afly+bay-\-cctfi=O ...... (22), 

which is satisfied by (21) on condition (13). proving the proposi- 
tion. 

NOTE. This problem was solved by Professor William Hoover, A. M., 
Ph. D , Professor of Mathematics and "Astronomy in the Ohio University, 
Athens, Ohio, who is one of the leading mathematicians in the United 
States, and whose biography follows. 



GEOMETRY. 



BIOGRAPHY. 

PROF, WILLIAM HOOVER, A. M., PH. D. 



Professor Hoover was born in the village of Smithville, Wayne county, 
Ohio, October 17, 1850, and is the oldest of a family of seven children. 
Both parents are living in the village where he was born, still enjoying 
good health. 

Up to the age of fifteen he attended the public schools, and for two or 
three years after, a local academy. Owing to needy circumstances he was 
obliged to work for his living quite early, and almost permanently closed 
attendance at any kind of school at eighteen years of age, sometime before 
which, going into a store in the county seat, as clerk. Nothing could have 
been farther from his taste than this work, having been thoroughly in love 
with study and books long before. After spending two or three years in 
this way, he went to teaching, about the year 1869, and he has been regularly 
engaged in his favorite profession to the present day. 

He attended Wittenberg College and Oberlin College one term each, a 
thing having very little bearing on his education. He studied no mathe- 
matics at either place, excepting a little descriptive astronomy at the latter. 

After teaching three winters of country school, with indifferent success, 
he was chosen, in 1871, a teacher in the Bellefontaine, Ohio, High School, 
serving one year, when he was given a place in the public schools of South 
Bend, Ind. Remaining there two years, he was invited to return to Belle- 
fontaine as superintendent of schools. He afterwards served in the same 
capacity in Wapakoneta, O., two years, and as principal of the second dis- 
trict school of Dayton, O. In 1883, he was elected professor of mathemat- 
ics and astronomy in the Ohio University, Athens, Ohio, where he is still 
in service. 

Through all his career of teaching, Professor Hoover has been an inces- 
sant student, devoting himself largely to original investigations in mathe- 
matics. Although his pretentions in other lines are very modest, he is em- 
inently proficient in literature, language, and history. Before going into 
college work he had collected a good library. He is indebted to no one for 
any attainments made in the more advanced of these lines, but by indefati- 
gable energy and perseverance he has made himself the cultured, classic, and 
renowned scholar he is. 

He has always been a thorough teacher, aiming to lead pupils to a mas- 
tery of subjects under consideration. His habits of mind and preparation 
for the work show him specially adapted to his present position, where he 
has met great success. He studies methods of teaching mathematics, 
which in the higher parts is supposed to be dry and uninteresting. He sets 
the example of enthusiasm as a teacher, and rarely fails to impress upon 
the minds of his students the immense and varied applications of mathe- 
matics. He is kind and patient in the class-room and is held in the highest 
esteem by his students. He is ever ready to aid the patient student inquir- 
ing after truth. It seems to be a characteristic of eminent mathematicians 
that they desire t'o help others to the same heights to which they them- 
selves haVe climbed. This was true of Prof. Seitz; it is true of Dr. Martin; 
and it is true of Prof. Hoover. 

In 1879, Wooster University conferred upon Prof. Hoover the degree of 
Master of Arts, and, in 1886, the degree of Doctor of Philosophy cum laude, 
he submitting a thesis on Cometary Perturbations. In 1889, he was 
elected a member of the London Mathematical Society and is the only 
man in his state enjoying this honor. In 1890, he was elected a member of 
the New York Mathematical Society. He has been a member of the Asso- 



404 FINKEL'S SOLUTION BOOK. 

ciation for the Advancement of Science for several years. Papers accepted 
bv the association at the meetings at Cleveland, Ohio, and at Washington, 
D. C., have been presented on "The Preliminary Orbit of the Ninth Comet 
of 1886," and "On the Mean Logarithmic Distance of Pairs of Points in 
Two Intersecting Lines." He is in charge of the correspondence work in 
mathematics in the Chautauqua College of Liberal Arts and of the mathe- 
matical classes in the summer school at Lake Chautauqua, the principal of 
which is the distinguished Dr. William R. Harper, president of the new 
Chicago University. The selection of Professor Hoover for this latter po- 
sition is of the greatest credit, as his work is brought into comparison with 
some of the best done anywhere. 

He is a critical readei and student of the best American and European 
writers, and besides, is a frequent contributor to various mathematical jour- 
nals, the principal of which are School Visitor, Mathematical Messenger, 
Mathematical Magazine, Mathematical Visitor, Analyst, Annals of Math- 
ematics, American Mathematical Monthly, and Educational Times, of Lon- 
don, England. 

His sU'le is concise and his aim is elegance in form of expression of 
mathematical thought. While greatly interested in the various branches of 
pure mathematics, he is specially interested in the applications to the ad- 
vanced departments of Astronomy, Mechanics, and the Physical Sciences 
such as Heat, Optics, Electricity, and Magnetism. The "electives" of- 
fered in the advanced work for students in his University are among the 
best mathematics pursued an> where in this country. 

He is an active member of the Presbyterian church and greatly interested 
in every branch of church work. He has been an elder for a number of 
years and was chosen a delegate to the General Assembly meeting at Port- 
land, Oregon, in May, 1892, serving the church in this capacity with fidel- 
ity and intelligence. In this biography of Professor Hoover, there is a val- 
uable lesson to be learned. It is this: Energy and perseverance will bring 
a sure reward to earnest effort. We see how the clerk in a county seat 
store, in embarrassing circumstances and unknown to the world of thinkers, 
became the well known Professor of Mathematics and Astronomy in one of 
the leading Institutions of learning in the State of Ohio. "Not to know 
him argues yourself unknown." 



THE NINE-POINT CIRCLE. 

Proposition. If a circle be described about the pedal trian- 
gle of any triangle, it will pass through the middle points of the lines 
drawn from the orthocenter to the vertices of the triangle, and 
through the middle points of the sides of the triangle, in all, through 
nine points. (2) The center of the nine-point circle is the mid- 
dle point of the line joining the orthocenter and the center of the 
circumscribing circle of the triangle. (8) The radius of the nine- 
point circle is half the radius of the circumscribing circle of the 
triangle. (4) The centroid of the triangle also lies on the line 
joining the orthocenter and the center of the circumscribing circle 
of the triangle, and divides it in the ratio of 2\ 1< (5) The sides of 
the pedal triangle intersect the sides of the given triangle in the 
radical axis of the circumscribing and nine-point circles. (6) The 
nine-point circle touches the inscribed and escribed circles of the 
triangle. 





BENJAMIN PEIRCE. 






GEOMETRY. 405 

The Pedal Triangle is a triangle formed by joining the 
feet of the perpendiculars drawn from the vertices of a triangle 
to the opposite sides. 

The Orthocenter is the point of intersection of these per- 
pendiculars. 

Medial Lines, or Medians, are lines drawn from the 
vertices of a triangle to the middle point of the opposite sides. 

The Centroid is the point of intersection of the medians. 

The Radical Axis of two circles is the locus of the points 
whose powers with respect to the two circles are equal. 

Demonstration. Let^47?Cbe an) triangle, AD, BF, and 
CE the perpendiculars from the vertices to the opposite sides of 
the triangle. O is the. orthocenter. Join the points F, E, and 
D. Then FED is the pedal triangle. About this triangle, de- 
scribe the circle FEHDK. It will then pass through the mid- 
dle points Z, N) and R of the lines, OA, OB, and OC, and the 
middle points H, G, and A" of the sides AB, BC, and A C, in all, 
through nine points. 

Since the angles AFO and AEO of the quadrilateral are both 
right angles a circle may be described about it. For the same 
reason circles may be described about the quadrilaterals EBDO 
andODCF. Draw the lines 7^7? and RG. Now the angles 
FRE and FDE are equal, being measured by half the same 
arc FE. But FDE equals 1EDL, because AD bisects the 
angle EDF. .'. FRO equals %FDL. Both being measured by 
the same arc OF, and FRO being two times FDL, FRO is an 
angle at the center; therefore, since OC is the diameter of the 
circle circumscribing FODC, R is the middle point of OC. In 
like manner it may be proved that OB and OA are bisected in 
the points TV and JL respectively. Draw the line RG. The 
angles RGC and RGB are equal to two right angles. Also the 
angles R GB and RED are equal to two right angles, because 
they are opposite angles of a quadrilateral inscribed in a circle. 
Therefore R GC is equal to RED. But RED is equal to 
OBD, because both are measured by half the arc OD. .-. The 
angle RGC equals the angle OBD, and consequently the line 
RG is parallel to the line OB. But, since RG bisects OC in R 
and is parallel to OB, it bisects B C in G. In like manner, it may 
be shown that AB and A C are bisected by the nine-point circle 
in the points H and A" respectively. Hence, the circle passes, in 
all, through nine points. Q. E. D. 

(2.) Draw the perpendiculars GP, KP, and HP from the 
-middle points of the sides of the triangle. They all meet in a 
common point P which is the center of the circumscribing circle 
of the triangle. With P as a center and radius equal to PB, 



406 



FINKEL'S SOLUTION BOOK 



describe the circumscribing circle. Draw the perpendiculars SY~ 
SJ, and SZ to the middle points of the chords FK, EH, and 




FIG. 6. 

DG. These all meet in the same point S, which is the center of 
the nine-point circle. In the trapezoid PHEO, since SJ bi- 
sects EH in J and is parallel to PH, it bisects OP in S. Hence, 
the center of the nine-point circle is the middle point of the line 
joining the orthocenter and center of the circumscribing circle. 

Q. E. D. 

(3.) Draw the lines KN and PB. Since the angle KFN is 
a right angle,, the line KN is a diameter of the nine-point cir- 
cle. KP=%BO=BN. Since KP and BN are equal and 
parallel, KPBN is a parallelogram, and consequently KNBP. 
.-. SN\fiP. But SJV is the radius of the nine-point circle 
and BP is the radius of the circumscribing circle of the triangle. 
Hence, the radius of the nine-point circle is half the radius of 
the circumscribing circle. Q. E, D. 

(4.) Draw the medial lines BK, AG, and CH. Draw the 
line KH. Now the triangles KPH a\\& BOC are similai he- 



GEOMETRY. 



40T 






cause the sides of the one are respectively parallel to the sides of 
the other, and the line UK is half the line BC, because //and 
Jfare the middle points of the sides AB and A C. .'.BO=ZKP. 
The triangles KPQ and BOQ are similar, because the angles of 
one are respectively equal to the angles of the other. Then we have 
KP:KQ\:BO:BQ or KP'.BO: :KQ : BQ. But BO=>IKP. 
.-. BQ=ZKQ. .'. Q is the centroid and divides the line joining 
orthocenter and the center of the circumscribing circle in the 
ratio of 2:1. Q.E. D. 

Hence the line joining the centers of the circumscribing and 
nine-point circles is divided harmonically in the ratio of 2 : 1 by 
the centroid and orthocenter of the triangles. These two points 
are therefore centers of similitude of the circumscribing and 
nine-point circles. .-. Any line drawn through either of these 
points is divided by the circumferences in the ratio of 2 : 1. 

(5. ) Produce FE till it meets B C in P / . Since two opposite an- 
gles of the quadrilateral BEFC are equal to two right angles, a 
circle may be circumscribed about it. Then we have P / E. P / F 
=P'B. P' C ; therefore the tangents from P' to the circles are 
equal. Q. E. D. 

(6.) Let Obe the orthocenter, and /and (Hhe centers of the in- 
scribed and circumscribed circles. Produce AI to bisect the arc 
BC'in T. Bisect AO in L, and 
join GL, cutting A Tin S. The 
nine-point circle passes through 
Gj D, and L, and D is a right an- 
gle. Hence, GL is a diameter, 
and is therefore =R=QA. There- 
fore GL and QA are parallel. But 
QA=QT, therefore GS=GT= 
CTsm$A=2fisin^A. Also ST 
=2GScos8, 6 being the angle GST 
GTS. N being the center of 
the nine-point circle, its radius 
NG=%R ; and r being the radius 
of the inscribed circle, it is required 
to show that NI=NGr. NI* 
=SN* -f- 5/ 2 2 SN. S/cos 6. Sub- 
stitute SN=R GS; SI=TI 
ST=2fism%A2GScos0; and G 
proposition. If J be the center of the escribed circle touch 
ing B C, r l its radius, it is shown in a similar way that 
JVJ=NG+r l . 




FIG. 7. 

to prove the 



408 FINKEIv'S SOLUTION BOOK. 

THE THREE FAMOUS GEOMETRICAL PROBLEMS OF 

ANTIQUITY. 

The limits of this work forbid our carrying the discussion of 
elementary geometry further. We have given merely an outline 
of how the subject may be studied by the student and presented 
by the teacher and that is our chief aim in this work. But before 
leaving the subject, it will be of interest to briefly speak of three 
famous problems in geometry, problems that have profoundly 
interested the mathematicians from the time of Plato down to 
the present time. These problems have been referred to before 
in this book so that, at this point, we shall only bring them to- 
gether and speak of them more explicitly. The problems re- 
ferred to are, 

(i) The Duplication of the Cube; (ii) the Trisection 
of an Angle; (iii) the Quadrature of the Circle. 

The first of these problems means to find the edge of a cube 
whose volume shall be twice that of a given cube; the second 
means to divide any given angle into three equal parts; and 
the third means to find the side of a square whose area shall be 
equal to that of a given circle. 

As has been said, constructions in pure geometry or Euclidean 
Geometry admit of the use of an ungraduated ruler and a pair 
of compasses only. With this restriction, all three problems 
are insoluble. This is the important point to be observed. The 
problems are only impossible, because we are limited in the use 
of our instruments to a straight edge, or ungraduated ruler, and 
a pair of compasses. In this way many problems may be made 
impossible. For example, it is impossible to go across the At- 
lantic Ocean from Boston to Liverpool on a bicycle, but with a 
steamship the trip is made very easily. So too, if other instru- 
ments are used our three problems are easily solved. The solu- 
tions of the first and second problems are implicitly involved in 
the Galois theory as presented to-day in treatises on higher 
algebra. The impossibility of the solution of the third was 
demonstrated in 1882 by Lindemann. 

The first two problems may be reduced to one, viz., that of 
finding two means between two given extremes. In the first prob- 
lem, if we let a be the edge of given cube and x that of the re- 

?uired cube, then we must have x* = 2a 3 , or a : JT = x : y = y : 20. 
n the second, if a is the sine of the given angle, and x the sine 

of one-third the angle ; then 4^ = $x a, or I : $x = $x ' y 



The problem of the duplication of the cube was known in 
ancient times as the Delian problem, in consequence of a legend 
that the Delians had consulted Plato on the subject. It is as- 
serted by Philoponus, that the Athenians in 430 B. C. were suf- 



PROBLEMS. 



409- 



fering from the plague of eruptive typhoid fever and in order to 
stop it consulted the oracle at Delos as to how it might be done. 
Appolo replied that they must double the size of the altar of 
Minerva which was in the form of a cube. This to the un-' 
learned suppliant, was an easy task, and a new altar having 
each of its edges double that of the old one was constructed,, 
in consequence of which the volume was increased eight-fold. 
This so enraged the god that he made the pestilence worse than 
before, and informed a fresh deputation that it was useless to 
trifle with him as the new alter must be a cube and have a. 
volume exactly double that of the old one. Suspecting a mystery,, 
the Athenians applied to Plato who referred them to the geom- 
etricians. In an Arab work, it is related that Plato replied to 
them, saying, Ye have been neglectful of the science of geometry 
and, therefore, hath God chastised you, since geometry is the 
most sublime of all the sciences. 

Many solutions of this problem have been given, one of which 
is given on page 234, by means of the Cissoid. We here give 
another by means of the parabola. 




Let y* ax, be the equation of the parabola whose axis coin- 
cides with axis of abscissas and x* 2#j/, the equation of the 
parabola whose axis coincides with the axis of ordinates. Solv- 
ing these two equations, we find j 3 2# 3 , that is, PM S = 2a 3 . 
Hence, if a is the edge of the given cube PMis the edge of the 
required cube. 

To trisect an angle, we proceed as follows: 

Let AOB be the given angle. With O as center and any 
radius, describe a circle, ABD. Draw the secant BDCso that 
DC shall be equal to the radius OB. (This is impossible unless 
a graduated ruler is used.) Then draw OD. Then angle BC^ 
= angle AOB. For angle DCO = angle DOC. Why? Angle 
BDO = 2 angle DCO. Why? Angle DCO + angle CBO = 
angle BOA. Why? .'. Angle AOB = 3 angle BCO. Why? 



410 FINKEL'S SOLUTION BOOK. 

The following elegant solution is due Claraut 




Let AOB be the given angle. With O as center and any 
radius describe a circle. Draw AB and trisect it in //"and K, so 
that AH= HK = KB. Bisect the angle AOB by OC t cutting 
AB in L. Then AH = ^HL. With focus A, vertex H, and 
directrix OC, describe a hyperbola. Let the branch of this 
hyperbola which passes through H cut the circle in P. Draw 
PM perpendicular to OC and produce it tq cut the circle in Q. 
Then by the focus and directrix property, we have AP : PM = 
AH:HL = 2:\. .'.AP=2 PM= PQ. Hence, by symmetry, 
AP = PQ = QR. Hence, AOP = POQ = QOR. 

The Quadrature of the Circle is effected by the Quadratrix. 
See page 238. 

For a very full treatment of these problems, see Klein's 
Famous Problems of Elementary Geometry, translated from the 
German by Professors Beman and Smith, also see Mathematical 
Recreations and Problems, by W. W. R. Ball. 




JAMES JOSEPH SYLVESTER, U,. D., F. R. S. 



PROPOSITIONS. 411 

PROPOSITIONS. 

1. The lines which join the middle points of adjacent sides of any quad- 
rilateral, form a parallelogram. 

2. Two medians of a triangle are equal; prove (without assuming that 
they trisect each other) that the triangle isisoscles. 

3. In an indefinite straight line AB find a point equally distant from 
two given points which are not both on AB. 

When does this problem not admit of solution? 
Construct a right triangle having given: 

4. The hypotenuse and the difference of the sides. 

5. The perimeter and an acute angle. 

6. The difference of the sides and an acute angle. 

7. Construct a triangle having given the medians. 

8. Construct a triangle, having given the base, the vertical angle, and(l) 
the sum or (2) the difference of the sides. 

9. Describe a circle which shall touch a given circle at a given point, and 
also touch a given straight line. 

10. Describe a circle which shall pass through two given points and 
be tangent to a given line. 

11. Find the point inside a given triangle at which the sides subtend 
equal angles. 

12. Describe a circle which shall be tangent to two intersecting straight 
lines and passing through a given point. 

13. Divide a triangle in two equal parts by a line perpendicular to a 
side. 

14. Inscribe in a triangle, a rectangle similar to a given rectangle. 

15. Construct an equilateral triangle equivalent to a given square. 

16- Trisect a triangle by straight lines drawn from a given point in one 
of its sides 

17. Draw through a given point a straight line, so that the part of it in- 
tercepted between a given straight line and a given circle may be divided 
at the given point in a given ratio. 

18. Construct a circle equivalent to the sum of three given circles. 

19. Find the locus of a point such that the sum of its distances from three 
given planes is equal to a given straight line. 

20. Construct a sphere tangent to three given spheres and passing 
througn a given point. 

21. Draw a circle tangent to three given circles. 

NOTE. This proposition is known as the Taction Problem, It was pro- 
posed and solved by Apollonius, of Pergse, A. D. 200. His solution was indi- 
rect, reducing the problem to ever simpler and simpler problems. It was 
lost for centuries, but was restored by Vieta. The first direct solution was 
given by Gergonne, 1813. An elegant solution of this problem is given by 
Prof. E. B. Seitz, School Visitor, Vol. IV, p. 61. 

22. Construct a sphere tangent to four given spheres. 
NOTE. This problem was first solved by Fermat (16011665). 

23. The perpendicular from the center of gravity of a tetrahedron to any 
plane without the tetrahedron is one-fourth of the sum of the perpendic- 
nlarsfrom the vertices to the same plane. 



412 FINKEL'S SOLUTION BOOK. 

1. Define: a segment of a circle, four proportional magnitudes, two 
similar polygons, the projection of a segment of a straight line on another 
straight line. 

2. The sum of all the plane angles about a point 'is four right angles. 

3. The locus of all points equally distant from two fixed points is the 
s.raight line that bisects the line joining the two points, at right angles. 

4. A straight line that is perpendicular to a radius at its extremity is. 
tangent to the circle ; and conversely. 

5. Two polygons that are similar to a third polygon aie similar to- 
each other. 

6. If two triangles have an angle of the one equal to an angle of the 
other, their areas are to each other as the rectangles of the sides includ- 
ing those angles. 

7. The ratio of the circumference of a circle to its diameter is the 
same for all circles. 

8. Find the side of the largest square that can be cut from a tree whose 
circumference is 14 feet. 

Cornell University Entrance Examination, 1899. 

[Proofs by limits are not, in general, satisfactory.] 

1. Define : a plane, a straight line perpendicular to a plane, a straight 
line parallel to a plane, two parallel planes, a diedral angle, the plane 
angle of a diedral. 

2. The sum of the face angles of a convex polyderal angle is less 
than four right angles. 

3. The sections of a prismatic surface made by two parallel planes are 
equal polygons. 

4. The frustum of a triangular pyramid is equal in volume to three 
pyramids, whose common altitude is the altitude of the frustum, and 
whose bases are the two bases of the frustum and a mean proportional 
between them. 

5. To draw a plane tangent to a cylindar with circle base. 

6. If two angles of a spherical triangle be equal, the opposite sides 
are equal. 

7. The lateral area of a cone of revolution is half the product of the 
perimeter of its base and its slant height. 

8. A cylindrical pail is 6 inches deep and 7 inches in diameter : find 
how much water it holds, and how much tin it takes to make it. 

Cornell University Entrance Examination, 



1. Define a straight line (preferably without using the ideas of dis- 
tance or direction). Also define: equal, greater, limit of a variable, 
length of a curve. 

2. If two triangles have two sides of one equal to two sides of the 
other, and the included angles of the first greater than that of the 
second, prove that the third side of the first is greater than the third 
side of the second. Also prove the converse of this theorem. 

3. Similar triangles (and similar polygons) are to each other as the 
squares on homologous sides. 

4. Construct a triangle, being given the lengths of the three per- 
dendiculars from the vertices on the opposite sides. 

5. Compute the side of a regular pentagon inscribed in a circle 
whose radius is given. 



PROBLEMS. 413 

6. Two straight lines in space have one and but one common per- 
pendicular, and it is the shortest line that can be drawn from one to 
the other. 

7. Compute the volume of a regular octahedron whose edge is two 
units. 

8. Show how to find the radius of a given sphere by means of meas- 
urements made on the surface. 

9. Prove that the volume of a cone is equal to the area of the base 
multiplied by one-third of the altitude. Also state without proof the 
chain of propositions which lead up to this theorem. 

10. Find the locus of a point in space the ratio of whose distances 
from two given points is equal to a given constant. 

Cornell University Scholarship Examination, 1899, 
Time, 3 hours. 

1. State and prove the theorem of Menelaus and its inverse. 

2. Prove: Circles described on any three chords from one point of 
a circle as diameters, have their other three points of intersection co- 
straight. 

3. Explain the Peaucellier Cell. 

4. State and prove the dual theorem of: The pole of any straight 
through a point is on the polar of the point. 

5. Prove: The diagonal triangle of a cyclic quadrangle is self-conju- 
gate (its own reciprocal polar). 

6. (a) Explain what is meant by a cross ratio of a range of four 
points. 

(fr) If (PQRS) =3, what are the other distinct cross ratios of the 
same range, and what are their magnitudes? 

(c) Deduce the distinct values of the cross ratios of a harmonic 
range. 

7. Prove Pascal's theorem concerning a cyclic hexagon. 

() State the corollaries as the number of sides is diminished. 

8. What is the radical axis of two circles? 

(b) Prove: The difference between the squares of the tangents 

from any point to two circles equals twice the rectangle of the center sect 

of the circles and the perpendicular sect from the point to the radical axis. 

Examination in Halsted's Modern Geometry. The University 

of Texas, 1894. Time, 3 hours. 

1. From the common notion "solid" as a starting point, define surface, 
line, point, plane, straight line. 

(b) Define angle, and point out the angles determined by a bi- 
radial. 

(r) What is meant by the statement that two magnitudes are 
equivalent? that one magnitude is greater than another? 

(d) Define the terms, multiple, submultiple, fraction, ratio. 

(e) What is the direct meaning of the statement that two series 
of magnitudes are proportional? State the simplest criteria of propor- 
tionality between two series of magnitudes in which to every one of either 
series there corresponds one of the other series. Apply the test to two 
such series where it is fulfilled ; and again where one criterion fails. 

(/) When are two figures perspective? when similar? 

2. Discuss the problem: To describe a circle tangent to three given 
intersecting straights, not all through the same point. Also, the analogous 
problem in a sphere. 



414 FINKEL'S SOLUTION BOOK. 

3. (a) State the conditions of congruence of two plane triangles. 

(b) State the conditions of similarity of two plane triangles. 

(c) State the conditions of congruence of two spherical triangles. 

4. (a) Investigate the form of the quadrilateral made by joining the 
mid points of consecutive sides of a quadrilateral. 

(fr) Its relative size. 

5. Divide a sect internally and externally in a given ratio. 

6. (a) Prove: If four sects are proportional the rectangle of the 
extremes is equivalent to the rectangle of the means. 

(b) State the inverse. 

7. (a) Prove the Pythagorean theorem without using any other con- 
cerning equivalence of figures. 

(b) Prove: The altitude to the hypothenuse is a mean propor- 
tional between the segments of the hypothenuse. 

8. (a) When is one spherical triangle A'B'C the polar of another. 
ABC? 

(fr) Prove: The sides of a spherical triangle intersect the cor- 
responding sides of its polar on the polar of their orthocenter. 

Examination in Hoisted' s Elementary Synthetic Geometry. The 
University of Texas, 1894. Time, 3 hours. 



ALGEBRA. 415 



ALGEBRA 



1 . Algebra is that branch of mathematics which treats of 
the general theory of operations with numbers, or quantities. 

The operations of ordinary abstract arithmetic are a particular case of 
algebra. Thus, algebra is sometimes called generalized arithmetic and in 
turn arithmetic is sometimes called specialized algebra. 

2. An Operation, in mathematics, is the act of passing 
from one number to another, the second number having a defi- 
nite relation to the first. 

3. An Operator, in mathematics, is a letter or symbol 
designating the operation to be performed. 

Thus, -{-, , X -*- Vi or & x > and f are operators. 
dx l 

4. The Fundamental Laws of Algebra are the Com- 
mutative Law; the Associative I/aw; and the Distribu- 
tive L/aw. 

The Commutative L/aw states, that additions and subtrac- 
tions may be performed in any order ; also the factors of a product 
may be taken in any order. 

The Associative If aw states, that the terms of an expres- 
sion may be grouped in any order. Thus, a+6 c+d e=(a+b) 
c+(d-e)=a+(b-c)+(d-e)=a+b-(cd}-e. 

Also the factors of a product may be grouped in any order. 

The Distributive I/aw states, that the product of a com- 
pound expression by a single factor is the algebraic sum of the 
partial products of each term of the compound expression by that 
factor. 

Thus, (a+b)c=ac+bc. 

For a very excellent treatment of these laws, the reader is . 
referred to Chrystal's Algebra, Part I. 

Note. The establishment of these thtee great laws was left forthe present century, 
the chief contributors thereto being De Mor 
were 
Pierce 
from those of ordinary algebra. 

The student who would become proficient in mathematics 
should make himself familiar with ordinary algebra, for it is the 
basis of all advanced mathematical subjects. For example, in 



416 



FINKEL'S SOLUTION BOOK. 



analytical geometry, the subject matter is geometry while the 
language is algebraic; in the calculus, the subject matter may 
be physics, astronomy, or political economy while the language 
is algebraic. We shall solve a few problems in algebra, leaving 
the student to gain a thorough knowledge of the subject by a 
study of such, works as Chrystal's Algebra, 2 vols. 

I. An estate was divided among three persons in such a 
way that the share of the first was three times that of the sec- 
ond, and the share of the second twice that of the third. The 
first received $900 more than the third. How much did each 
receive? [From Hall and Knight's College Algebra, p. 69, 
prob 40.} 

1. I^et ^r=the number of dollars in the share of the 

third person. Then 

2. 2^=the number of dollars in the share of the sec- 

ond, and 

3. 6^r=3X2^r=the number of dollars in the share of 

the first. 

4. Qx .#=5.*=the number of dollars the first received 

more than the second. 

II. -I 5. 900= the number of dollars the first received 
more than the second. 

6. .'. 5^=900, 

7. x=\ of 900=180, the number of dollars in the 

share of the third, 

8. 2^=360, the number of dollars in the share of 

the second, 

9. 6^=1080, the number of dollars in the share of 

the first. 

( $180=share of the third, 
III./. <' $360= " " " second, and 
($1080= " " " first. 

I. The length of a room exceeds its breadth by 8 feet; if 
each had been increased by 2 feet, the area would have been in- 
creased by 60 square feet; find the original dimensions of the 
room. [ From Hall and Knight's College Algebra, p. 69, prob. jj.] 

f 1. L,et ^=the number of feet in the breadth of the 
room. Then 

2. ^-j-8=the number of feet in the length, and 

3. (x+8)x=x' 2 -\-8x=the number of square feet in the 

area. 

4. jr+2=the conditional number of feet in the 

breadth, and 

5. ^-j-2+8=^-|-10=the conditional number of feet in 

the length. Then 




ARTHUR CAYLEY. 



ALGEBRA. 



417 



II. { 6. (#4-2) (#-1-10)=: # 2 -|-12#-{-20:= the conditional num- 
ber of square feet in the area. 

7. (# 2 +12#4-20)-(# 2 +8#)=4#+20=the number of 

square feet in the increase in the area, 

8. 60=the number of square feet in increase in area. 

9. .'. 4#4-20rr:60, 

10. 4#=40, by subtracting 20 from both sides. 

11. #= of 40=10, the number of feet in the breadth, 

and 
^12. #4-8=: 18, the number of feet in the length. 



Ill 



. 1 10 feet=the breadth, and 
' ' ' (18 feet=the length. 



I. A takes 3 hours longer than B to walk 30 miles; but if 
'he doubles his pace, he takes 2 hours less time than B; find 
their rates of walking. [From Hall and Knight's College Alge- 
' 



II. 



1. 


L,et #=A's rate in miles per hour, and 




2. 


B . s M ,i 




3. 


30 
Then =number of hours it takes A to 


travel 30 




X 






miles. 




4. 


30 
= number of hours it takes B to travel 


30 miles. 




y 






30 30 




5. 


.'. =3, by the first condition of the 


problem. 


6. 


2.r:=A's conditional rate in miles per hour. 




30 15 




7. 


Then -^= ^number of hours it takes A to travel 




LX X 






30 miles. 






30 15 





8. 


/. =2, by the second condition of 


the prob- 




lem. 






15 15 3 




9. 


_LC/ .LtJ ZJ /(*\ 

=-p-, from (5). 




10. 


= 3J, by adding (8) and (9). 






y 






1 7 




11. 






12. 


30 
/. j/=-=- =4^=number of miles B travels 


per hour. 


13. 


15 15 3 
_ =-^-, by substituting for_y in (9). 




14. 


15 7 3 

T~ 2 ~ 2 ' 





418 



FINKEIv'S SOLUTION BOOK. 



-' 

17. .'. -r=3rrnumber of miles A travels per hou*. 



Ill 



. ( 3 miles=A's rate per hour, and 
' ' ' { 4| miles=B's rate per hour. 



I. In a mile race A gives B a start of 44 yards and beats 
him 51 seconds. In the second trial A gives B a start of 1 min- 
ute and 15 seconds, and is beaten by 88 yards. Find the rate of 
each in miles per hour. \Todhunter* s Algebra, p. ioj, prob. 23; 
Wentworttts Complete Algebra, p. 179, prob. 55.] 



II. 



(i. 

2. 
3. 

4. 
5. 

6. 
7. 

8. 

9. 
10. 
11. 
12. 
13. 
14. 
15. 

16. 


lyet .z=A's rate in yards per second. 
j^=rB's rate in yards per second. 
1 mile=1760 yards. 


n the first 

in second 

I) by 1760. 
5) by 1672. 
[); whence 

per hour, 
in (3) and 

cond. 


1760-44 1716 


y y 

trial. 
. 1716 1760 


y x 


1760-88 1672 


trial. 
. 1760 1672 


= t& . . . . \4). 

39 1 51 


40j/ x 1760 
20 1 75 f 


l % *581 67 V 


760y ~~ 33440 ' ^ v subtracting (3) from (' 

44 

y=<r yards, B's rate per second. 

3600 44 


=-oo, by substituting the value of y 

changing the signs. 

88 
. * . X=-^F yards= A's rate in yards per se 



ALGEBRA. 



-IT -m ., 3600 88 ,, 

17. .'.12 miles= .. _ X A s rate m 



419 
per hour. 



Ill I ^ miles:=B's rate per hour. 
' ' * 1 12 miles=A's rate per hour. 



per 

I. THE QUADRATIC EQUATION. 
5. ax?+6x+c=Q, is the general quadratic equation. 






x= 



In the solution of exercises involving quadratic equations, 
students should be thoroughly grounded in the method of com- 
pleting the square, and this method should not be superseded by 
the Hindoo Method nor the Method of Factoring, though this 
latter method should receive due consideration. When the 
method is thoroughly impressed upon the mind of the student 
he should be encouraged to solve examples by merely substitut- 
ing in the general formulae above. 

Thus, find the values of x satisfying the equation 

0. Here, a=2, =5, and *= 33. 



Then 



x , ==. 



5+T/25-4-2 33 __5+17 
2-2 4 



Xo= 



5-i/25-4-2 33 ~ 







2-2 



I. Find the price of eggs per score when 10 more in 
cents' worth lowers the price 31 cents per hundred. [ Went 
worttts Complete Algebra, p. 216, prob. <?.] 

1. Let x= price per score. 

2. -|j=price per egg. 



420 



FINKEL'S SOLUTION BOOK. 



II. 



3. 5* price of 100 eggs. 

4. 62^-^--= =number of eggs in 62J cents' 






worth. 



c 1250 , 1A 1250+10* , .- 1A 

5. - +10= ^number if 10 more be 
x x 

added. 

C01 1250+10* 25* , ., 1A 

6- 62^ 3 = 50Q+4;r =price of each egg, if 10 



* 



7, 



more be added to 62| cents' worth. 

625* 

price of 100 eggs. 



125+* 



.'. 5* 



625* 



, 

9. x 



125+* 
25* 3125 



* . r r 

clearing of fractions, transpos- 



\ 4 

ing, combining, and dividing through by the 
coefficient of * 2 . 
1A , 25 . 625 50625 , , . 

10. * "T" jr +~5T" = ~~gi > b y completing the square. 

25 225 

11. * = ^ , by extracting the vSquare root. 

12. *=31J ; or -25. 



III. .'. 31Jc.=price of the eggs per score. The negative 
value is inadmissible in an arithmetical sense. 

x*+y =11 (1)) 

\ Find the values of * and r. [From Schuy- 

x +y*= 7 (2)J 
ler>s Complete Algebra, p. 368, prob. 4.] 

1. * 2 9=2 y (3), by transposing in (1). 

2. * 3=4 y 2 (4), by transposing in (2). 

3. --3== 



II. < 



4. 



6. K 2 ~ 



=-^, or 



y 



*+3 
1 



(6), 
=4 



y transposing. 
2 1 



*+3 4 (*+3) 2 *+3 4 (*+3) 2 
completing the square in (6). 
1 1 



(7), by 



root of (7). 



)> by extractin S the S( l uare 



ALGEBRA. 



421 



8. .'. y=2, by transposing 




in (8). 



^7, by substituting value of y in (2). 



III. \ -* 



x and j have three other values in addition to those found. 
For a number of different solutions giving the four values of x 
andjy, see The American Mathematical Monthly. 



Find x, y, and z. 



For a solution of this example, see The Mathematical Mag- 
azine, edited by Dr. Artemas Martin, Washington, D. C. 

I. Find two numbers whose product is equal to the differ- 
ence of their squares, and the sum of whose squares is equal to 
the difference of their cubes. [Ray's Higher Algebra, p. 230, 
prob. p.] 

Let ^=greater number, 

and j/=less number. 

xy=x*-y* (1). 

x*+y*=x*y* (2). 

Let x=ay, then 

ay 2 =a 2 y 2 y* (4), by substituting the value of x 

in (1). 

a 2 a 1 (5), by dividing (4) byjj/ 2 , and arranging, 
whence a=|-(l=fcV 5 , by completing the square of 
(5), and extracting the square root, and trans- 
posing. 



II. < 






9. y = -I^ = 

2(1V 5) 
in (2). 

10. x=ay=m- 



substituting the value of a 



5) (JV 5) = 



5). 



III. 



and 



II. INDETERMINATE FORMS. 



6. The symbol, 0, is defined by the equation a a=Q. It is 
not used to denote nothing, but is used to denote the absence 
of quantity. 

All operations upon this symbol are impossible. Thus, 0X5, 



422 FINKEL'S SOLUTION BOOK. 

5-K), 5, are all impossible operations. Standing apart from the 
conditions imposed upon the quantities from which arises by 
certain limitations, the operations above indicated are absolutely 
meaningless. But such indicated operations, and many others 
of the same nature, do very frequently occur in mathematical 
investigations, and when they do thus arise they must be inter- 
preted in such a way as to conform to the fundamental laws of 
mathematics. 

In conformity to these laws, OX=<2XO=0; a+Q=Q+a=a; 
0^=0. 

The symbol, GO , is used to represent a quantity that is larger 
than any assignable quantity, however large. 

What meaning shall be attached to the following indicated 
operation, #-K)? It is impossible to perform this operation. 
Suppose we divide a by h. This is possible, provided a and h 

are real numbers, and is indicated thus,. Now what happens 

to the value of the fraction , if we conceive h to diminish 

indefinitely? We know that as the denominator of a fraction 
decreases, the value of the fraction increases. Hence, if the 
value of the denominator becomes very small, the value of the 
fraction becomes very large. If the denominator becomes less 
than any assignable quantity, the value of the fraction becomes 
larger than any assignable quantity. All this is concisely and 

accurately stated as follows: I -7- I GO , or I I =00 . 

h=Q L k J L h J A= 

Hence, the inaccurate though common form, a-s-O, must be 
interpreted in the light of the above explanation and, therefore,, 
a-*-0=oo , briefly though inaccurately expressed. 

tf=l, for all finite values of a\ but is indeterminate if a is GO . 
O a =0, for all finite values of a\ but is infinite if a is infinite, 
fl-s-oo =0, for all finite values of a; but is indeterminate if a is 
infinite. is indeterminate, oo GO , co -5-00 , CH-0, OXoo , and 
I 00 are all indeterminate. But when these forms occur in any 
mathematical investigation, they usually have a determinate 
value. 

^a^-i ~l 

Thus, . =. -?-=a+x I . =%a. Here, -TT- 2a. 

a~X \x=a \ x = a 



All the other forms may be reduced to the form -^T. 

a a a a . a OXa 

ThUS, 00-00= _-_=_=_; 0X00 =0 X ^ -- -g- = -g- 



ALGEBRA. 42S 

a b a v 0X0 ... 

=" x = ; log ' (1 )= 



Since the log. (1) is indeterminate, the quantity 1 s0 is indeter- 
minate. 

It is important that the student masters the meaning of these 
forms, as they occur very frequently in the higher mathematics. 
For example, the Differential Calculus rests largely upon the 

proper interpretation of -rr. 



r . 

1. Find the limiting value of ^ r-r, when x=2 

x' 2 HbH-16 

I x-2x-3~l 



-_i 

(*-2)(.r-8)_= 2 " ^-S^ =2 "-6 

^2-1-2^ 

2. Find the limiting value of ~ - when x=Q and when 



_ _ 

jf 



3. Find the limiting value of 8 _-|o . ^ when jr=2 and 
when JT=QO . 



00 



/y/ ^J y g. 

4. Find the limiting value of = - ^=z when x a* 

r =F ^ 



424 FINKEIv'S SOLUTION BOOK. 

Let x=a+k, where h approaches as a limit. Then 



.- A 2+ etc. 



-% o 

EXAMPLES. 

i 

X 



1. Find the limiting value of ., when jr=0. 



r V^ a 

2. Find the limiting value of - - when x=a. 

xa 

]_ _ ^"^ 

3. Find the limiting value of -- g-= when ^=1. 

1 v x 



4. Find the limiting value of 2 when x=~L. 
.5. a "-^1 =what? 

=what? 



x a 



. x 
7 =what? 



8 . _ =what 



III. ARITHMETICAL FALLACIES. 
7. First Fallacy. Assume that a=b. Then 



.' . ba-\-b 

.: b=Zb. 
.'. 1=2, 



ALGEBRA. 425 

8. Second Fallacy. Let a and b be two unequal numbers, 
and let c be their arithmetical mean. Then 



.'. acbc 
.'. a=b. 

Since we assumed that a and b were unequal, where is the fallacy 
in our reasoning? 

9. Third Fallacy. We have ( 1) 2 =1. Taking logarithms, 

2 log. (-l)=log. 1=0 
.-.log. (-1)=0 
.-. l=<?o 
.'. 1=1, since ^=1. 

10. Fourth Fallacy. We know that 

log. (l+^)=^-i^a+i^8- ---- 

If x\, the resulting series is convergent; hence 

log. 2=l-J+J-i+i-i4^-i-H- .... 
2 log. 2-2-l+f-^+f-J+|-i+-|- ..... 

Taking those terms together which have a common denomina- 
tor, we obtain 

2 log. 2=l+i-J+i+ 



-log. 2. 
.-.2=1. 

11. Fifth Fallacy. We can write V 11 "! = V ~\ in the 
form 

= J 

V^T_ vT~ 
' v~i iFT 

.-. -1=1, 

12.' 5V.rM Fallacy, r The mathematical theory of probability 
leads to various paradoxes; of these, one specimen follows: Sup- 
pose three coins to be thrown up and the fact whether each comes 
down head or tail to be noticed. The probability that all the 
coins come down head is (J)*, that is, ; similarly, the probabil- 
ity that all three coins come down tail is ^; hence, the probabil- 
ity that all come down alike, that is, either all of them heads or 
all of them tails, is J. But, of three coins, thus thrown up, at. 



426 FINKEVS SOLUTION BOOK. 

least two must come down alike. The probability that the third 
comes down head is -J and the probability that it comes down 
tail is ^. Thus the probability that it comes down the same as 
the other two is ^. Hence, the probability that all the coins 
come down alike is J. 

~IV. PROBABILITY. 

13. Definition. If an event can happen in a ways and 
fail in b ways, and each of these ways is equally likely, the 

probability, or the chance of its happening is - ^-r-, and the 
chance of its failing is rr . 

d-TO 

That is, the probability of an event happening is the number 
of favorable ways the event can happen divided by the total 
number of ways it can happen, and the probability of its failing 
is the number of ways the event can fail divided by the total 
number of ways it can happen. 

For example, if in a lottery there are 6 prizes and 23 blanks, 
the chance that a person holding 1 ticket will win a prize is T 6 F , 
and his chance of not winning is ff . 

14. The reason for the above definition may be made clear 
by the following considerations : 

If an event can happen in a ways and fail to happen in b 
ways, and all these ways are equally likely to occur, we can 
assert that the chance of its happening is to its chance of failing 
as a to b. Thus if the chance of its happening is represented 
by ka, where k is an undetermined constant, then the chance of 
its failing is kb. 

.'. Chance of happening + chance of failing = k(a-\-b). Now 
the event is certain to happen or to fail; therefore, the sum of 
the chances of happening and failing must represent certainty. 
If, therefore, we agree to take certainty as our unit, we have 



=, or = 
~ 

.*. the chance that the event will happen is -^rr an( i the 

chance the event will not happen is r-r. 

15. The subject of probability, from the mathematical point 
of view, is a very difficult one. That it is a very important sub- 
ject, no one will deny after having read Jevons's Principles of 
Science, 2 vols., in the first volume of which he has given con- 
siderable attention to its treatment. Our space is too limited to 
give here more than a passing reference to the subject. Those 
who desire to study the subject thoroughly, should read Tod- 



ALGEBRA. 



427 



hunter's History of the Theory of Probability; LaPlace's Theorie 
Analytique des Probabilities, 1812 (the most exhaustive treat- 
ment of the subject ever written); and Whitworth's Choice and 
Chance. The last named book has a large number of problems 
worked out in full. 

EXAMPLES. 

1. I. What is the chance of throwing a number greater 
than 4 with an ordinary die whose faces are numbered from 1 
to 6? 

6=number of ways the die can fall. 

2r=number of ways the die can fall so as to give a 

number greater than 4, viz., 5 and 6. 
3. .*. == the required chance, by definition. 



II. 



III. /. the required chance is J. 

2. I. Find the chance of throwing at least one ace in a 
single throw with two dice. 

1. 6=number of ways one of the dice may fall. 

2. 6X6=36=:number of ways the two dice may fall, 

since with each of the six ways the first may 
fall, there are six ways in which the second 
may fall. 

3. 5=number of ways one die may be thrown with- 

out the ace coming up. 

4. 25=5X5=number of ways the two dice can be 

thrown without either of them being ace. 

5. .*. ffr^the chance of not throwing an ace. 

6. .'. 1 f=iJ=the chance of throwing at least one 

ace. 

III. .*. the chance of throwing at least one ace is . 

3. I. If four coins are tossed find the chance that there 
should be two heads and two tails. 



II. 



J^chance of head or tail with one coin. 
^==(jj*=chanice of all heads, tossing 4 coins. 
-j^=(^) 4 =chance of all tails, tossing 4 coins. 
i~ f^i(i) 3 chance of 1 head and 3 tails. 
J=iX(J)*(J)=chance of 1 tail and 3 heads. 



r 1. 
2. 
3. 
4. 
5. 



. 
6 - t^?i) 2 (i) 2 -chance of 2 heads and 2 tails. 



II 



III. .'. the chance of throwing 2 heads and 2 tails is f. 

4. I. A bag contains 5 white, 7 black, and 4 red balls; find 
the chance that three balls drawn at random are all white. 



428 



FINKEIv'S SOLUTION BOOK. 



. 16X15X14=number of ways 3 objects can be se- 
lected from 16 objects. 

. 5X4X3=number of ways 3 objects can be selected 
, from 5 objects, which is the number of ways the 

3 white balls may be selected from the 5 white 
balls. 
5X4X3 ( . 

' ' 16X15X14 =A=the 
III. .'. the required chance is 



PROBLEMS. 

1. State as a theorem the fact implied in the following equa- 
tions : 9 2 7 2 = 4-8; 5 2 - - 3 2 = 4.4; 93 2 - - 9i 2 = 4-92; 
32 j2 _ 4.2 Prove it, and then express the theorem in its 
most general terms. 

2. How find the highest common factor of two polynomials 
that cannot be readily factored? Prove your answer. Illustrate 
by finding the H. C. F. of 

6x* + 7^ 2 $x and 15^* + 3I* 3 + io^r 2 . 

3. A cistern can be filled in 4 hours by two pipes running 
together, and in 6| hours by one of the pipes alone. In how 
many hours can the other pipe fill the same cistern? 

4. If V a and V b are surds, prove that V a V b cannot be 
a rational number. 



5 . Simplify 




+ y 



2 TT 



1/2 



Check your work by substituting x 4 and y = I, both in the 
given expression and in the simplified form, and comparing 
results. 
6. Given the two simultaneous equations 

find all the pairs of values of x and y that satisfy them. 

Cornell University Entrance Examination, 1899. 

T. 



i. Resolve 



3* 



into partial fractions. 



2. At an election there are 4 candidates, and 3 members to 



PROBLEMS, 429 

be elected, and an elector may vote for any number of candidates 
not greater than the number to be elected. In how many ways 
may an elector vote ? 

2 

3. Find, by using logarithms, the value of V41.72 X (.054) 3 . 

4. Show that for any two quantities, the square of their geo- 
metric mean is equal to the product of their arithmetic and har- 
monic means. 

5. Draw the graph of the function x z + ,r 2 + x 100; and 
find the root between 4 and 5, correct to three places of decimals, 
of the equation ,r 3 -}- .r 2 -f- x 100 = o. 

6. If h, k, are the roots of ax* + bx + c o, find the value 
, 1 1 

f ^+^" 

7. The square of x varies as the cube of y, and x = 3 when 

1 

y 4. Find the value of y when x ~~rw- 

V o 

Cornell University Entrance Examination, 1899. 

I. (a) Simplify the expression 

Z6e _ 26c _ 

J+~^ b J+7 
i H~ i i 



1 



Show how the symmetry of this expression may be made to serve 
as a partial check upon the result. 

(b) As x becomes more and more nearly equal to 2, what 

value does the fraction approach? What is the 

value of this fraction when x 2? 

2. The sum of the three digits of a number is 9; the digit 
in hundreds' place equals J of the number composed of the two 
other digits, and the digit in units' place equals -J of the number 
composed of the two other digits ; what is the number ? 

3. Prove that : (o) x* + y n is exactly divisible by x + y if n 
is any odd, positive integer whatever. 

(b) If a + b is constant, then a-b is greatest when a = 6. 

4. (a) State what seem to you to be the chief differences 
between algebra and arithmetic. 

(b) Define negative number, subtraction, and multiplica- 
tion, and show how, from your definition, the following rules may 
be deduced: (i) "change the sign of the subtrahend and pro- 
ceed as in addition"; (2) give the product the positive or the 



430 FINKEL'S SOLUTION BOOK. 

negative sign according as the two factors have like or unlike 
signs." 

5. Given the equation ax 2 + bxy + cy z = o in which a, b, 
and c are real, the ratio x\y being unknown. Find the sum 
and also the product of the roots (i. e. of the values of the ratio 
x:y) of this equation. If a approaches o relatively to b and c, 
what happens to these roots? For what relative values of a, b, 
and c are the two roots equal? One twice the other? Both 
imaginary ? T. 

Cornell University Scholarship Examination, 1899. 

1. Two men, A and B, had a money-box containing $210, 
from which each drew a certain sum daily ; this sum being fixed 
for each, but different for the two. After six weeks, the box 
was empty. Find the sum which each man drew daily from the 
box ; knowing that A alone would have emptied it five weeks 
earlier than B alone. 

Obtain two solutions, and interpret the negative answer. 

2. Solve the equation 

x + b 2a _., 2a /-, 2a ~ 6\ 

$a -X.-6 ' ~~~6~\ x b) 

3. Reduce to their lowest common denominator 

i i 
and 



- 6 4-3T 3 - 6x 2 - 4^ + 6 

and find, and reduce to their lowest terms, the difference and the 
quotient of these two fractions. 
4. Write out (x y) 9 . 

(4 Vb z 1\ 9 

o g -- \aVa.b T } ; reducing the an- 

swer to the simplest possible form, in which it is free from nega- 
tive and fractional exponents, and has only one radical sign. 
Harvard University Entrance Examination, 

1. Resolve into three factors (x 2 x) B 8. 

2. Find the greatest common factor of 

x* + 5* z + 6> and 3^.+ 7^ + 3^ + 2. 



Solve the equation V^r-4 + V^ 11- V2*9=0. 



Vr 
*J x 



4. Simplyfy 

x-y 



PROBIvEMS. 431 

5. Solve the simultaneous equations 

3O 2 + xy) 403;, x y 2. 

6. What is the geometrical mean between 

2x 3 and zx* + x* ^x 3 ? 

7. A and B start at the same time from the same point in the 
same direction. A goes at the uniform rate of 60 miles per day ; 
B goes 14 miles the first day, 16 miles the second day, 18 miles 
the third day and so on. At the end of 50 days who will be ahead, 
and by how much? 

Massachusetts Institute of Technology Entrance Ex- 
amination in Advanced Algebra, 1892. 

1. (a) Show that 

(w+i) (n + 2) (n + 3 ) n 

-_i =( n * + 6 n+ n). 
2 + 3 6 

(b) Find the algebraic expression which when divided 
by x 2 2X -{- i gives a quotient x 2 -\- 2x + i and a remainder 
x i. 

2. (a) Reduce to a common denominator (arranging the 
terms of the numerator according to ascending powers of x) 

ABC D 

- + - + -- h - 
x x+i O+l) 2 (*+i) 3 

( b ) Having given that : A = 2, B 2 A o, C 2B 
-+ 3 A= 3, D 2C + 38 =_i,_ 2 D + 3 c= ; find the 
values of B, C, D, and E. 

3. Solve for x and y, 2X 3^, + 14 = o, 53; ^x = 26. 

4. (a) Simplify j&+9xt-4r(x*)9-*r+x&-*r. 

(*) Multiply xn+ Jt +lbyx - a+ J +l . 

5. Solve for x (a) x^ -\- 2a? = $ax. Also 

(^)/ + 3^ = 4. 

6. Solve the simultaneous equations 



Also (&) (x + y) = 20, x* + y 2 = 2a 2 . 

Princeton University Entrance Examination, 1893. 

I. Write the factors of the following expressions: 
x^ (x 6) 2 and w 6 64W 6 . 



432 FINKEL'S SOLUTION BOOK. 

12 



2. Simplify x i 



12 

5 



3. A and B can do a piece of work in m days ; B and C can 
do it in n days ; C and D in p days ; and D and A in r days. In 
how many days can all working together do it ? 

4. Multiply x + y V z by y 2 ^x. 

5. Solve the equation y 2 + 2 (a -f 6) y = i8a<. 

6. Extract the square root of 



1 



7. Simplify. 

l+T 

8. Write the 6th term of (a 2&) J . 

Yale University Entrance Examination, 1893. 

1. (a) Solve the equation ax 2 -j- for -f- c = o. 

(fr) What relation must exist between its coefficients in 
order (i) that its roots may be imaginary, (2) that they may 
be real and equal, (3) that they may be real and unequal? 

(c) If the coefficient a diminishes without limit, what 
limits, if any, do the roots respectively approach ? 

2. Make the first members of the following equations perfect 
squares, without introducing fractions: 

2^2 _ ^ x 2j 3 ^2 _ g^ _ 4 

3. Solve completely the simultaneous equations x 2 -j- xy + y 2 
:= 19, x 2 xy + 3> 2 = 7, and group distinctly the corresponding 
values of x and y. 

4. Convert 3.14159 into a continued fraction, and obtain four 
convergents. What is the limit of the error in taking the fourth 
convergent for the value of the decimal ? 

5. (a>) Derive the formula for the number of permutations 
of m things taken n at a time. 

(b) From 10 different things in how many ways can a 
selection of 4 be made ? 



PROBLEMS. 433 

6. (a) Write equivalents for the following expressions: 
log 1; log ; log fl O, if a>l ; log a 2 - log 2 ; 



(b) Given the mantissa of log 10 257 = 0.40993, to find 

log 10 T/0257: 

7. Given log a N and log a b, to find log b N. 

Sheffield Scientific School of Yale University Entrance 
Examination in Advanced Algebra, 1892. 



434 



FINKEL'S SOLUTION BOOK. 



PROBABILITY PROBLEMS. 



I. If three pennies be piled up at random on a horizontal 
plane, what is the probability that the pile will not fall down? 

The pile will stand if the common center of gravity of the sec- 
ond and third coins falls on the surface of the first or bottom, 
coin. 

Let r be the radius of a penny ; then the center of the second 
coin may fall anywhere in a cir- 
cle whose radius is 2r and cen- 
ter the center of the surface of 
the first or bottom coin, and the 
center of the third coin may fall 
anywhere in a circle whose ra- 
dius is 2r and the center the 
center of the surface of the sec- 
ond coin. The number of posi- 
tions of the center of the second 
coin is therefore proportional to 
47Tr 2 , and for every one of these 
positions the center of the third 
coin can have 4?rr 2 ; hence the 
total number of positions of the ^ G ' s ' 

second and third coins is proportional to 167rV 4 . 

We must now determine in how many of these 167rV 4 posi- 
tions the pile will stand. 

Let A be the center of the first or bottom coin, and B the cen- 
ter of the second coin. Take AD=AB, and with center D 
and radius 2r describe the arc CHJ. If the center of the third 
coin is on the surface CFJH, the second and third coins will re- 
main on the first, since BN=NH, BT=TC, and the pile will 
not fall down. 

When AB is not greater than -Jr, the circle CHJ will not cut 
the surface of the second coin, and the pile will stand if the cen- 
ter of the third coin is anywhere on the second. 




Let AJ3=AD=x,S=surface CFJH, and /=the probability 
required ; then DB=%x, BG=J tr *~^ X ' 



, arc Cf=r cos" 1 



PROBABILITY PROBLEMS. 435 



, and 



1 1 



/.r / /''Sr 2 _ 4* 2> \ 

- / ( S nr* )xdx. lr* cos- 1 ( * r A }xdx, 

V/ J ^ rx y 



(K^2 _ 4v2x 1 
- 1 4?F- > )-^( 5r2 - 




NOTE. This solution is due to Artemas Martin, M. A., Ph. D., LL. D., 
member of the London Mathematical Society, member of the Edinburg 
Mathematical Society, member of the Mathematical Society of France, 
member of the New York Mathematical Society, member of the Philo- 
sophical Society of Washington and Fellow of the American Association 
for the Advancement of Science, Washington, D. C., who is one of the 
great peers of mathematical science. 



436 FINKEL'S SOLUTION BOOK. 



BIOGRAPHY. 

ARTEMAS MARTIN, M- A., PH. D., LL. D. 



This eminent mathematician was born in Steuben county, N. Y., Au- 
gust 3, 1835- Early, his parents moved to Venango county, Pa., where they 
lived for many years. Dr. Martin had no schooling in his early boyhood, 
except a little primary instruction; but by self-application and indefatiga- 
ble energy which have told the story of many a great man, he has become 
familiar to every mathematician and lover of science in every civilized 
country of the world 

He was never a pupil at school, except when quite small, until in his 
fourteenth year. He had learned to read and write at home, but knew 
nothing of Arithmetic. At fourteen he commenced the study of Arithme- 
tic, and after spending two winters in the district school, he commenced 
the study of Algebra. At seventeen, he studied Algebra, Geometry, Nat- 
ural Philosophy, and Chemistry in the Franklin Select School, walking 
two and one-half miles night and morning. Three years after, he spent 
two and one-half months in the Franklin Academy, studying Algebra and 
Trigonometry. This finished his schooling. He taught district schools 
four winters, but not in succession. He was raised on a farm, and worked 
at farming and gardening in the summer; chopped wood in the winter; 
and after the discovery of oil in Venango county, worked at drilling oil 
wells a part of his time, always devoting his "spare moments" to study. 

In the spring of 1869, the family moved to Erie county, Pa., where he re- 
sided until he entered the U. S. Coast Survey Office in 1885. While in 
Erie county, after 1871, he was engaged in market-gardening, which he 
carried on with great care and skill. He began his mathematical career 
when in his eighteenth year, by contributing solutions to the Pittsburg 
Almanac, soon after contributing problems to the "Riddler Column" of the 
Philadelphia Saturday Evening Post, and was one of the leading contribu- 
tors for twenty years. 

In the summer of 1864 he commenced contributing problems and solu- 
tions to Claris School Visitor, afterward the Schoolday Magazine, pub- 
lished in Philadelphia. In June, 1870, he took charge of the "Stairway De 
partment" as editor, the mathematical department of which he had con- 
ducted for some years before. He continued in charge as mathematial edi- 
tor till the magazine was sold to Scribner & Co., in the spring of 1875, at 
which time it was merged into ''6Y. Nicholas" 

In September, 1875, he was chosen editor of a department of higher 
mathematics in the Normal Monthly, published by Prof. Edward Brooks, 
Millersville, Pa., and held that position till the Monthly was discontinued 
in August, 1876. He published in the Normal Monthly a series of sixteen 
articles on the Diophantine Analysis. 

In June, 1877, Yale College conferred on him the honorary degree of 
Master of Arts (M. A.) In April, 1878, he was elected member of the Lon- 
don Mathematical Society. In June, 1882, Rutgers College conferred on 
him the honorary degree of Doctor of Philosophy (Ph. D.) March 7, 1884, 
he was elected a member of the Mathematical Society of France. In April, 
1885, he was elected a member of the Edinburgh Mathematical Society. 
June 10, 1885, Hillsdale College conferred on him the honorary degree of 
Doctor of Laws (LL. D.) February 27, 1886, he was elected a member of 
the Philosophical Society of Washington. In June, 1881, he was elected 
Professor of Mathematics of the Normal School at Warrensburg, Mo., but 
did not accept the position. November 14, 1885, Dr. Martin was appointed 



PROBABILITY PROBLEMS. 



437 



Librarian in the office of the U. S. Coast and Geodetic Survey. On August 
'26, 1890, he was elected a Fellow of the American Association for the Ad- 
vancement of Science. On April 3, 1891 he was elected a member of the 
New York Mathematical Society. 

All these honors have been worthily bestowed and the Colleges and So- 
cieties conferring them have done honor to themselves in recognizing the 
merits of one who has become such a power in the scientific world through 
his own efforts. 

He has contributed fine problems and solutions to the following journals 
of the United States: School Visitor, Analyst, Annals of Mathematics, 
Mathematical Monthly, Illinois Teacher, I oiva Instructor, National Edu- 
cator, Tates County Chronicle, Barnes' Educational Monthly, Wittenberger, 
Maine Farmers' Almanac, Mathematical Messenger, and Educational 
Notes and Queries, American Mathematical Monthly. Besides other con- 
tributions, he contributed thirteen articles 011 "Average " to the Mathemat- 
ical Department of the Wittenberger, edited by Prof. William Hoover. 
These are believed to be the first articles published on that subject in 
America. 

Dr. Martin has also contributed to the following English mathematical 
periodicals : Lady's, and Gentleman's Diary, Messenger of Mathematics, 
and The Educational Times and Reprint. 

The Reprint contains a large number of his solutions of difficult "Aver- 
age " and "Probability" problems, which are master-pieces of mathematical 
thought and skill, and they will be lasting monuments to his memory. His 
style is direct, clear and elegant. His solutions are neat, accurate and sim- 
ple. He has that rare faculty of presenting his solution in the simplest 
mathematical language, so that those who have mastered the elements of 
the various branches of mathematics are able to understand his reasoning. 

Dr. Martin is now ( 1899 ) editor of the Mathematical Magazine, and 
The Mathematical Visitor, two of the best mathematical periodicals pub- 
lished in America. These are handsomely arranged and profusely illustra- 
ted with very beautiful diagrams to the solutions, he doing the typesetting 
with his own hand. The typographical work of these journals is said to be 
the finest in America. The best mathematicians from all over the world 
contribute to these two journals. The Mathematical Visitor is devoted to 
Higher Mathematics, while The Mathematical Magazine is devoted to the 
solutions of problems of a more elementary nature. All solutions sent to 
Dr. Martin receive due credit, and if it is possible to find room for them the 
solutions are all published. He has thus encouraged many young aspirants 
to higher fields of mental activity. He is always readv to aid any one who 
is laboring to bring success with his work. He is of a kind and noble dis- 
position and his generous nature is in full sympathy with every diligent 
student who is rising to planes of honor and distinction by self application 
and against adverse circumstances. 

Dr. Martin has a large and valuable mathematical library containing 
many rare and interesting works. His collection of Amerioan arithmetics 
and algebras is one of the largest private collections of the, kind in this 
country. 



438 FINKEL'S SOLUTION BOOK. 



I. Find the average or mean distance 
from one corner. 

Taking the corner from which the mean distance is to be 
found for the origin of orthogonal co-ordinates, and one of the- 
sides of the square for the axis of abscissa, we have for the ele- 
ment of the surface dx dy, and since this element is at a distance: 

1 C a f* a 
from the origin, the average / / dxdy*J(x 2 -\-y%) 






.: Average=ia[V2+log(l+V2)]. 

NOTE. This solution is by Prof. J. W. F. Sheffer, Hagerstown, Md.,, 
whose name may be found attached to the solutions of many difficult prob- 
lems proposed in the leading mathematical journals of the United States. 
The above solution is taken from the Mathematical Messenger, published 
by G. H. Harville, Simsboro, La. 

I. All that is known concerning the veracities of two witnesses, 
A and B, is that B's statements are twice as reliable as A's.. 
What is the probability of the truth of the concurrent testimony 
of these two witnesses? 

Let #=the probability of the truth of any one of A's state- 
ments ; then 2#=the probability of any one of B's statements.. 
The event did occur if both witnesses tell the truth, the proba- 
bility ot which is xX^x=2x 2 . The event did not occur if both 
testify falsely, the probability of which is (1 #)x(l 2ff)=l 
3#-|-2tf 2 . Hence, the probability of the occurrence of the event 
on the supposition that x is known is 



x ) (I -2*)' 

Now, as the veracity of B can not exceed unity, the greatest 
value of* is found by putting 2#=1, which gives #=-J; hence, 
x can have any value from to -J-. 

Therefore, the probability in the problem is 

., , * 

P dx 



C 

Jo 



3) 2 +7* 

Let 8 3=y. Then x=$(y 3), dx=\dy\ the limits of> are 
1 and 3, and 






PROBABILITY PROBLEMS. 



439< 



NOTE. This solution is taken from the Mathematical Magazine, Vol. 
II, p. 122. The solution there given is credited to the author, Prof. William 
Hoover, and Prof. P. II. Philbrick. 

I. A cube is thrown into the air and a random shot fired 
through it; find the chance that shot passed through oppo- 
site faces. 

Let AH\)Q the cube. Through P ', a point in the face EFGH , 
draw MK parallel to HE, and draw PN perpendicular to HE* 
Now if PA represents the direction of 
the shot, it will pass through the face 
ABCD, if it strikes the face EFGH 
anwhere within HMPN. 




FIG. 9. 



and area HMPN=u. Then 
PK=sec 6 tan0, 7^T=tan 
sec0sec0, PM=\ sec0tan0, 
tan#, ?te( 1 sec#tan0) ( 1 tan# ) ,the 
area of the projection of HMPN on a 
plane perpendicular to PA=ucos6cos(f>, 
and that of EFGH=co$8cos(f). 

Since we are to consider all possible 
directions of the shot with respect to the cube, the points of in- 
tersection of PA with the surface of a sphere whose center is A, 
and radius unity, must be uniformly distributed. An element 
of the surface of this sphere is co&cfid 6 d<{). By reason of the 
symmetry of the cube, the required chance is obtained by finding 
the number of ways the shot can pass through the opposite faces 
EFGH and ABCD between the limits 6=0, and 0=^ and 
0=0 and 0=tan~ 1 (cos#)=0', and the number of ways it can 
pass through the face .S/^G^/between the limits 0=0 and #=-j-;r,. 
and 0=0 and <fc=^rt ; and then dividing the former by the latter. 
Hence, the chance required is 



p 

r 

Jo 



= f /?* f 

nj Jo 



6*2 tan" 1 



tan 0) 






NOTE. This solution is due to Professor Enoch Beery Seitz, member 
of the London Mathematical Society, and late Professor of Mathematics 
in the North Missouri State Normal School, Kirksville, Mo. 



440 FINKEL'S SOLUTION BOOK. 

BIOGRAPHY. 



PROF. E. B. SEITZ, M. L. M. S. 

Professor Seitz, a distinguished mathematician of his day, was 
born in Fairfield Co., p., Aug. 24, 1846, and died at Kirksville, Mo., Oct. 8, 
1883. His father, Daniel Seitz, was born in Rockingham Co., Va., Dec. 17, 
1791, and was twice married. His first wife's name was Elizabeth Hite, of 
Fairfield Co., O., by whom he had eleven children. His second wife's name 
was Catharine Beery, born in the same county, Apr. 11, 1808, whom he 
marr-ed Apr. 15, 1832. This woman was blessed by four sons and three 
daughters. Mr. Seitz followed the occupation of a farmer and was an in- 
dustrious and substantial citizen. He died near Lancaster, O., Oct. 14, 
1864, in his seventy-third, year ; having been a resident of Fairfield Co. for 
sixty-three years. 

Professor Seitz, the third son by his father's second marriage, passed 
his boyhood on a farm, and like most men who have become noted, had 
only the advantages of a common school education. Possessing a great 
thirst for learning, he applied himself diligently to his books in private and 
became a very fine scholar in the English branches, especially excelling in 
arithmetic. It was told the author, by his nephew, Mr. Huddle, that when 
Professor Seitz was in the field with a team, he would solve problems while 
the horses rested. Often he would go to the house and get in the garret 
where he had a few algebras upon which he would satiate his intellectual 
appetite. This was very annoying to his father who did not see the future 
greatness of his son, and many and severe were the floggings he received 
for going to his favorite retreat to gain a victory over some difficult prob- 
lem upon which he had been studying while following the plow. Though 
the way seemed obstructed, he completed algebra at the age of fifteen, with- 
out an instructor. He chose teaching as his profession which he followed 
with gratifying success until his death. He took a mathematical course in 
the Ohio Wesleyan University in 1870. In 1872, he was elected one of the 
teachers in the Greenville High School, which position he held till 1879. 
On the 24th of June, 1875, he married Miss Anna E. Kerlin, one of Darke 
county's most refined ladies. In 1879, he was elected to the chair of mathe- 
matics in the Missouri State Normal School, Kirksville, Mo., which posi- 
tion he held till death called him from the confines of earth, ere his star of 
fame had reached the zenith of its glory. He was stricken by that "demon 
of death," typhoid fever, and passed the mysterious shades, to be numbered 
with the silent majority, on the 8th of October, 1883. On the llth of 
March, 1880, he was elected a member of the London Mathematical So- 
ciety, being the fifth American so honored. 

He began his mathematical career in 1872, by contributing solutions to 
the problems proposed in the ''Stairway" department of the Schoolday 
Magazine, conducted by Artemas Martin. His masterly and original solu- 
tions of difficult Average and Probability problems, soon attracted universal 
attention among mathematicians. Dr. Martin being desirous to know what 
works he had treating on that difficult subject, was greatly surprised to 
learn that he had no works upon the subject, but had learned what he knew 
about that difficult department of mathematical science by studying the 
problems and solutions in the Schoolday Magazine. Afterwards, he con- 
tributed to the Analyst, the Mathematical Visitor, the Mathematical Maga- 
zine, the School Visitor, and the Educational Times, of London, Eng. 

In all of these journals. Professor Seitz was second to none, as his 
logical and classical solutions of Average and Probability problems, rising 
as so many monuments to his untiring patience and indomitable energy and 
perseverance will attest. His name first appeared as a contributor to the 
Educational Times in Vol. XVIII., of Reprint, 1873. From that time until 
his death the Reprint is adorned with some of the finest product of his 
mighty intellect. 



BIOGRAPHY. 441 

On page 21, Vol. II., he has given the above solution. This problem 
had been proposed in 1864 by the great English mathematician, Prof. Wool- 
house, who solved it with great labor. It was said by an eminent mathe- 
matician of that day that the task of writing out a copy of that solution 
was worth more than the book in which it was published. 

No other mathematician seemed to have the courage to investigate the 
problem after Prof. Woolhouse gave his solution to the world, till Profes- 
sor Seitz took it up and demontsrated it so elegantly in half a page of or- 
dinary type, that he fairly astonished the mathematicians of both Europe 
and America. Prof. Woolhouse was the best English authority on Proba- 
bilities even before Professor Seitz was born. 

It was the solution of this problem that won for Professor Seitz the 
acknowledgment of his superior ability to solve difficult Probability prob- 
lems over any other living man in the world. 

In studying his solutions, one is struck with the simplicity to which he 
has reduced the solutions of some of the most intricate problems. When 
he had grasped a problem in its entirety, he had mastered all problems of 
that class. He would so vary the condition in thinking of one special 
problem and in effecting a solution that he had generalized all similar cases, 
so exhaustive was his analyses. Behind the words he saw all the ideas 
represented. These he translated into symbols, and then he handled the 
symbols with a facility that has rarely been surpassed. 

What he might have accomplished in his maturer years, had he turned 
his splendid powers to investigations in higher and more fruitful fields of 
mathematics, no man may say. The solving of problems alone is not a 
high form of mathematical research. While problem solving is very bene- 
ficial and essential at first, yet, if one confines himself to that sort of 
work exclusively, it becomes a waste of time. 

He was a man of the most singularly blameless life; his disposition 
was amiable; his manner gentle and unobtrusive; and his decision, when 
circumstances demanded it, was prompt and firm as the rocks. He did 
nothing from impulse; he carefully considered his course and came to 
conclusions which his conscience approved; and when his decision was 
made, it was unalterable. 

Professor Seitz was not only a good mathematician, but he was also 
proficient in other branches of knowledge. His mind was cast in a large 
mold. ''Being devout in heart as well as great in intellect, 'signs and quan- 
tities were to him but symbols of God's eternal truth' and 'he looked 
through nature up to nature's God.' Professor Seitz, in the very appro- 
priate words of Dr. Peabody regarding Benjamin Pierce, Professor of 
Mathematics and Astronomy in Harvard University, 'saw things precisely 
as they are seen by the infinite mind. He held the scales and compasses 
with which eternal wisdom built the earth, and meted out the heavens. As 
a mathematician, he was adored by his friends with awe. As a man, he 
was a Christian in the whole aim and tenor of life.' " 

Professor Seitz did not gain his knowledge from books, for his library 
consisted of only a few books and periodicals. He gained such a profound 
insight in the subtle relations of numbers by close application with which 
he was particularly gifted. He was not a mathematical genius, that is, as 
ususally understood, one who is born with mathematical powers fully de- 
veloped. But he was a genius in that he was especially gifted with the 
power to concentrate his mind upon any subject he wished to investigate. 
This happy faculty of concentrating all his powers of mind upon one topic 
to the exclusion of all others, and viewing it from all sides, enabled him to 
proceed with certainty where others would become confused and disheart- 
ened. Thread by thread and step by step, he took up and followed out 
long lines of thought and arrived at correct conclusions. The darker and 
more subtle the question appeared to the average mind, the more eagerly 
he investigated it. No conditions were so complicated as to discourage 
him. His logic was overwhelming. 



442 FINKEL'S SOLUTION BOOK. 



BIOGRAPHY. 



RENE' DESCARTES. 

Rene Descartes, the first of the modern school of mathematicians, 
was born at La Haye, a small town on the right bank of the Creuse and 
about midway between Tours and Poitiers, on March 31st, 1596, and 
died at Stockholm, on February llth, 1650. "The house is still shown 
where he was born, and a metairie about three miles off still retains the 
name of Les Cartes. His family on both sides was of Poitevin descent 
and had its headquarters in the neighboring town of Chatterault, where 
his grandfather had been a physician. His father, Joachim Descartes, 
purchased a commission as counsellor in the Parlement Rennes and thus 
introduced the family into that demi-noblesse of the robe of which, in 
stately isolation between the bourgeoisie and the high nobility, main- 
tained a lofty rank in the hierarchy of France. For one-half of each year 
required for residence the elder Descartes removed, with his wife, 
Jeanne Brochard, to Rennes. Three children, all of whom first saw 
the light at La Haye, sprang from the union, a son, who afterwards 
succeeded to his father in the Parlement, a daughter who married a M. 
du Crevis, and a second son, Rene. His mother, who had been ailing be- 
forehand, never recovered from her third confinement; and the mother- 
less infant was intrusted to a nurse, whose care Descartes in after years 
remembered by a small pension."* 

Descartes, who early showed an inquisitive mind, was called by his 
father, "my philosopher." At the age of eight, Descartes was sent to 
the school of La Fleche, which Henry IV had lately founded and en- 
dowed for the Jesuits, and here he continued from 1604 to 1612. Of 
the education here given, of the equality maintained among the pupils, 
and of their free intercourse, he spoke at a later period in terms of 
high praise. Descartes himself enjoyed exceptional privileges. His 
feeble health excused him from the morning duties, and thus early he 
acquired the habit of matutinal reflection in bed, which clung to him 
throughout life. When he visited Pascal in 1647, he told him that the 
only way to do good work in mathematics and to preserve his health 
was never to allow any one to make him get up in the morning before 
he felt inclined to do so. Even at this period he had begun to distrust 
the authority of tradition and his teachers. 

Two years before leaving school (1610) he was selected as one of 
twenty-four gentlemen who went forth to receive the heart of the mur- 
dered king as it was borne to its Besting place at La Fleche. During 
the winter of 1612, he completed his preparations for the world by les- 
sons in horsemanship and fencing; and then in the spring of 1613 he 
started for Paris to be introduced to the world of fashion. Fortunately 
the spirit of dissipation did not carry him very far, the worst being a 
passion for gaming. Here through the medium of the Jesuits he made 
the acquaintance of Mydorge, one of the foremost mathematicians of 
France, and renewed his schoolboy friendship with Father Mersenne, and 
together with them he devoted the two years of 1615 and 1616 to the study 
of mathematics. 

"The withdrawal of Mersenne in 1614 to a post in the provinces was 
the signal for Descartes to abandon social life and shut himself up for 
nearly two years in a secluded house of the Faubourg St. Germain. 
Accident, however, betrayed the secret of his retirement; he was com- 
pelled to leave his mathematical investigations and to take a part in 
entertainments, where the only thing that chimed in with his theorizing 

*Britannica Encyclopedia, Ninth Edition. 



BIOGRAPHY. 443 

reveries was the music. The scenes of horror and intrigue which 
marked the struggle for supremacy between the various leaders who 
aimed at guiding the politics of France made France no fit place for a 
student and held out little honorable prospect for a soldier. Accordingly, 
in May, 1617, Descartes, now twenty-one years of age, set out for the 
Netherlands, and took service in the army of Prince Maurice of Orange, 
one of the greatest generals of the age, who had been engaged for some 
time in a war with the Spanish forces in Belgium. At Breda, he enlisted 
as a volunteer, and the first and only pay which he accepted he kept as a 
curiosity through life. There was a lull in the war; and the Nether- 
lands were distracted by the quarrels of Gomarists and Arminians. Dur- 
ing the leisure thus arising, Descartes one day, as he roved through 
Breda, had his attention drawn to a placard in the Dutch tongue; 
and as the language of which he never became perfectly master, was 
then strange to him, he asked a bystander to interpret it in either French 
or Latin. The stranger, who happened to be Isaac Beeckman, principal 
of the College of Dort, offered with some surprise to do so into Latin, 
if the inquirer would bring him a solution of the problem for the 
advertisement was one of those challenges which the mathematicians of 
the age, in the spirit of the tournament of chivalry, were accustomed to 
throw down to all comers, daring them to discover a geometrical mystery 
known as they fancied to themselves alone. Descartes promised and ful- 
filled ; and a friendship grew up between him and Beeckman broken 
only by the literary dishonesty of the latter, who in later years took 
credit for the novelty contained in a small essay on music (Compendium 
Musicae) which Descartes wrote at this period and intrusted to 
Beeckman."* 

The unexpected test of his mathematical attainments afforded by the 
solution of the problem referred to, its solution costing him only a few 
liours study, made the uncongenial army life distasteful to him, but under 
family influence and tradition, he remained a soldier, and was pursuaded 
at the commencement of the thirty years' war to volunteer under Count 
de Bucquoy in the army of Bavaria. The winter of 1619, spent in 

Siarters at Neuburg on the Danube, was the critical period in his life, 
ere, in his warm room (dans un poele), he indulged those meditations 
which afterwards led to the Discours de la Methode (Discourse of 
Method). It was here that, on the eve of St. Martin's day, November 
10, 1619, he "was filled with enthusiasm, and discovered the foundations 
of a marvelous science." 

He retired to rest with anxious thoughts of his future career, which 
haunted him through the night in three dreams, that left deep impres- 
sions on his mind. "Next day," he says, "I began to understand the 
first principles of my marvelous discovery." Thus the date of his philo- 
sophical conversion is fixed to a day. This day marks the birth of 
modern mathematics. His discovery, viz., the cooperation of ancient 
geometry and algebra, is epoch-making in the history of mathematics. 

It is frequently stated that Descartes was the first to apply algebra 
to geometry. This statement is not true, for Vieta and others had done 
this before him, and even the Arabs sometimes used algebra in connec- 
tion with geometry. "The new step that Descartes did take was the in- 
troduction into geometry of an analytical method based on the notion 
of variables and constants, which enabled him to represent curves by 
algebraic equations. In the Greek geometry, the idea of motion was 
wanting, but with Descartes it became a very fruitful conception. By 
him a point was determined in position by its distances from two fixed 
lines or axes. These distances varied with every change of position in 
the point. This geometric idea of co-ordinate representation together with 
the algebraic idea of two 'variables in one equation having an indefinite 
number of simultaneous values, furnished a method for the study of loci, 

"Encyclopedia Britannica, Ninth Edition. 



444 FINKEL'S SOLUTION BOOK. 

which is admirable for the generality of its solutions. Thus the entire 
conic sections of Appollonius is wrapped up and contained in a single 
equation of the second degree."f 

''Descartes found in mathematics, as did Kant and Comte, the type 
of all faultless thought; and he oroved his appreciation of his insighf 
by the invention of a new symbolic mechanism and artifice for the appli- 
cations of algebra to geometry (Analytic Geometry, as it is now calledj 
which, in a growing sense, let it be said, existed before him), and by 
his discoveries in the theory of equations, which were fundamental in 
their importance."* 

After a short sojourn in Paris, Descartes moved to Holland, then at 
the height of its power. There for twenty years he lived, giving up all 
his time to philosophy and mathematics. Science, he says, may be 
compared to a tree, metaphysics is the root, physics is the trunk, and 
the three chief branches are mechanics, medicine, and morals, these 
forming the three applications of our knowledge, namely, to the external 
world, to the human body, and to the conduct of life; and with these 
subjects alone his writings are concerned. 

He spent the time from 1629 to 1633 writing Le Monde, a work em- 
bodying an attempt to give a physical theory of the universe; but find- 
ing its publication likely to bring on him the hostility of the Church, 
and having no desire to pose as a martyr, he abandoned it. The in- 
complete manuscript was published in 1664. 

He then devoted himself to composing a treatise on universal science ; 
this was published at Leyden in 1637 under the title Discourse de la 
methode pour bien conduire sa raison et chercher la verite dans les 
sciences, and was accompanied with three appendices entitled La Diop- 
trique, Les Meleores, and La Geometric. It is from the last of these 
that the invention of analytical geometry dates. In 1641, he published a 
work called Meditations, in which he explained at some length his 
views of philosophy as sketched out in the Discourse. In 1644, he issued 
the Principia Philosophiae, the greater part of which was devoted to 
physical science especially the laws of motion and the theory of vor- 
tices. In his theory of vortices, he commences with a discussion of 
motion ; and then lays down ten laws of nature, of which the first 
two are almost identical with the first two as laid down by Newton. 
The remaining eight are inaccurate. He next proceeds to a discussion 
of the nature of matter which he regards uniform in kind though there 
are three forms of it. He assumes that the matter of the universe is in 
motion, that this motion is constant in amount, and that the motion 
results in a number of vortices. He states that the sun is the center of 
an immense whirlpool of this matter, in which the planets float and 
are swept round like straws in a whirlpool of water. 

Each planet is supposed to be the center of a secondary whirlpool by 
which its satellites are carried, and so on. All of these assumptions are 
arbitrary and unsupported by any investigation. It is a little strange 
that a man who began his philosophical reasonings by doubting all 
things and finally coming to cogito, ergo sum should have made assump- 
tions so groundless. 

While Descartes was a philosopher of a very high type, yet his fame 
will ever rest on his researches in mathematics. The first important 
problem solved by Descartes in his geometry is the problem of Pappus, 
viz. : "Given several straight lines in a plane, to find the locus of a point 
such that perpendiculars, or, more generally, straight lines at given angles, 
drawn from the point to the given lines, shall satisfy that the product of 
certain of them shall be in given ratio to the product of the rest." "The 
most important case of this problem is to find the locus of a point such 
that the product of the perpendiculars on m given lines be in a constant 

\Cajori' s History of Mathematics. 
*The Open Court, August, 1898. 



BIOGRAPHY. 445 

ratio to the product of the perpendiculars on n other given straight 
lines. The ancients had solved this geometrically for the case m = 1, 
n = 1, and the case w = 1, n = 2. Pappus had further stated that if 
w n 2, the locus was a conic, but he gave no proof. Descartes also 
failed to prove this by pure geometry, but he showed that the curve was 
represented by an equation of the second degree, that is, was a conic; 
subsequently Newton gave an elegant solution of the problem by pure 
geometry."* 

In algebra, Descartes expounded and illustrated the general methods 
of solving equations up to those of the fourth degree (and believed that 
his method could go beyond), stated the law which connects the posi- 
tive and negative roots of an equation with the change of signs in the 
consecutive terms, known as Descartes' Law of Signs, and introduced 
the method of indeterminate coefficients for the solution of equations. 

In appearance, Descartes was a small man with large head, project- 
ing brow, prominent nose, and black hair coming down to his eye- 
brows. His voice was feeble. Considering the range of his studies he 
was by no means widely read, had no use for Greek, as is shown by 
his disgust when he found that Queen Christina devoted some time each 
day to its study, and despised both learning and art unless something 
tangible could be extracted therefrom. In philosophy, he did not read 
much of the writings of others. In disposition, he was cold and selfish. 
He never married, and left no descendants, though he had one illegitimate 
daughter, Francine, who died in 1640, at the age of five. 

In 1649, through the instigation of his close personal friend, Chanut, 
he received an invitation to the Swedish court, and in September of that 
year he left Egmond for the north. Here, on the llth of February, 1650, 
he died of inflammation of the lungs brought about by too close devo- 
tion to the sick-room of his friend Chanut, who was dangerously ill 
with the same disease. By B. F. Finkel. From the American Mathe- 
matical Monthly. 

* Ball's Short Account of the History of Mathematics. 



446 FINKEIv'S SOLUTION BOOK. 



BIOGRAPHY. 



LEONHARD EULER. 

Leonhard Euler (oiler), one of the greatest and most prolific mathe- 
maticians that the world has produced, was born at Basel, Switzerland, 
on the 15th day of April, 1707, and died at St. Petersburg, Russia, No- 
vember the 18th (N. S.), 1783. Euler received his preliminary instruc- 
tion in mathematics from his father who had considerable attainments as 
a mathematician, and who was a Calvinistic* pastor of the village of 
Riechen, which is not far from Basel. He was then sent to the Univer- 
sity of Basel where' he studied mathematics under the direction of John 
Bernoulli, with whose two sons, Daniel and Nicholas, he formed a life- 
long friendship. Geometry soon became his favorite study. His genius 
for analytical science soon gained for him a high place in the esteem 
of his instructor, John Bernoulli, who was at the time one of the first 
mathematicians of Europe. Having taken his degree as Master of Arts 
in 1723, Euler afterwards applied himself, at his father's desire, to the 
study of theology and the Oriental languages, with the view of entering 
the ministry, but, with his father's consent, he returned to his favorite 
pursuit, the study of mathematics. At the same time, by the advice of 
the younger Bernoulli is, who had removed to St. Petersburg in 1725, 
he applied himself to the study of physiology, to which he made useful 
applications of his mathematical knowledge ; he also attended the lectures 
of the most eminent professors of Basel. While he was eagerly engaged 
in physiological researches, he composed a dissertation on the nature 
and propagation of sound. In his nineteenth year he also composed a 
dissertation in answer to a prize-question concerning the masting of ships, 
for which he received the second prize from the French Academy of 
Sciences. 

When his two close friends, Daniel and Nicholas Bernoulli, went to 
Russia, they induced Catherine I, in 1727, to invite Euler to St. Peters- 
burg, where Daniel, in 1733, was assigned to the chair of mathematics. 
Euler took up his residence in St. Petersburg, and was made an asso- 
ciate of the Academy of Sciences. In 1730 he became professor of physics, 
and in 1733 he succeeded his friend Daniel Bernoulli, who resigned on a 
plea of ill health. 

At the commencement of his astonishing career, he enriched the 
Academical collection with many memoirs, which excited a noble emula- 
tion between him and the Bernouillis, though this did not in" any way 
affect their friendship. It was at this time that he carried the integral 
calculus to a higher degree of perfection, invented the calculation of 
sines, reduced analytical operations to greater simplicity, and threw new 
light on nearly all parts of pure or abstract mathematics. In 1735, an 
astronomical problem proposed by the Academy, for the solution gi which 
several eminent mathematicians had demanded several months' time, 



*The Encyclopedia Britannica says Ruler's father was a Calvinistic minister, while 
W. W. R. Ball, in his History of Mathematics, says he was a Lutheran minister. Euler 
himself was a Calvinist in doctrine, as the following, which is his apology for prayer, 
indicates: "I remark, first, that when God established the course of the universe, and 
arranged all the events which must come to pass in it, he paid attention to all the cir- 
cumstances which should accompany each event ; and particularly to the dispositions, 
to the desires, and prayers of every intelligent being; and that the arrangement of all 
events was disposed in perfect harmony with all these circumstances. When, therefore, 
a man addresses God a prayer worthy of being heard it must not be imagined that such 
a prayer came not to the knowledge of God till the moment it was formed. That prayer 
was already heard from all eternity; and if the Father of Mercies deemed it worthy 
of being answered, he arranged the world expressly in favor of that prayer, so that the 
accomplishment should be a consequence of the natural course of events. It is thus 
that God answers the prayers of men without working a miracle." 



BIOGRAPHY. 447 

was solved by Euler in three days with the aid of improved methods of 
his own, but the effort threw him into a fever which endangered his life 
and deprived him of his right eye, his eyesight having been impaired 
by the severity of the climate. With still superior methods, this same 
problem was solved later by the illustrious German mathematician, 
Gauss. 

In 1741, at the request, or rather command, of Frederick the Great, 
he moved to Berlin, where he was made a member of the Academy of 
Sciences, and Professor of Mathematics. He enriched the last volume of 
the Melanges or Miscellanies of Berlin, with five memoirs, and these were 
followed, with astonishing rapidity, by a great number of important re- 
searches, which wefce scattered throughout the annual memoirs of the 
Prussian Academy. At the same time, he continued his philosophical 
contributions to the Academy of St. Petersburg, which granted him a 
pension in 1742. 

The respect in which he was held by the Russians was strikingly 
shown in 1760, when a farm he occupied near Charlottenburg happened 
to be pillaged by the invading Russian army. On its being ascertained 
that the farm belonged to Euler, the general immediately ordered com- 
pensation to be paid, and the Empress Elizabeth sent- an 'additional sum 
of four thousand crowns. The despotism of Anne I caused Euler, who 
was a very timid man, to shrink from public affairs, .and to devote all 
his time to science. After his call to Berlin, the Queen of Prussia who 
received him kindly, wondered how so distinguished a scholar should be 
so timid and reticent. Euler replied, "Madam, it is because I come from 
a country where, when one speaks, one is hanged." 

In 1766, Euler, with difficulty, obtained permission from the King of 
Prussia to return to St. Petersburg, to which he had been originally 
called by Catherine II. Soon after returning to St. Petersburg a cataract 
formed in his left eye, which ultimately -deprived him of sight, but this 
did not stop his wonderful literary productiveness, which continued for 
seventeen years until the day of his death. It was under these cir- 
cumstances that he dictated to his amanuensis, a tailor's apprentice who 
was absolutely devoid of mathematical knowledge, his Anleitung zur 
Algebra, or Elements of Algebra, 1770, a work which, though purely 
elementary, displays the mathematical genius of its author, and is still 
considered one of the best works of its class. Euler was one of the very 
few great mathematicians who did not deem it beneath the dignity of 
genius to give some attention to the recasting of elementary processes 
and the perfecting of elementary text-books, and it is not improbable 
that modern mathematics is as greatly indebted to him for his work 
along this line as for his original creative work. 

Another task to which he set himself soon after returning to St. 
Petersburg was the preparation of his Lettres a une Princesse d' Allemagne 
sur quelques sujects de Physique, (3 yols. 1768-72). These letters were 
written at the request of the princess of Anhalt-Dessau, and contain an 
admirably clear exposition of the principal facts of mechanics, optics, 
acoustics, and physical astronomy. Theory, however, is frequently un- 
soundly applied in it, and it is to be observed generally that Euler's 
strength lay rather in pure than in applied mathematics. In 1755, Euler 
had been elected a foreign member of the Academy of Sciences at Paris, 
and sometime afterwards the academical prize was adjudged to three of 
his memoirs Concerning the Inequalities in the Motions of the Planets. 
The two prize-problems proposed by the same Academy in 1770 and 
1772 were designed to obtain a more perfect theory of the moon's mo- 
tion. Euler, assisted by his eldest son, Johann Albert, was a competitor 
for these prizes and obtained both. In his second memoir, he reserved 
for further consideration the several inequalities of the moon's motion, 
which he could not determine in riis first theory on account 9f the 
complicated calculations in which the method he then employed had 



448 FINKEVS SOLUTION BOOK. 

engaged him. He afterward reviewed his whole theory with the assist- 
ance of his son and Krafft and Lexell, and pursued his researches until 
he had consiructed the new tables, which appeared with the great work 
in 1772. Instead of confining himself, as before, to the fruitless integra- 
tion of three differential equations of the second degree, which are fur- 
nished by mathematical principles, he reduced them to three ordinates 
which determine the place of the moon ; and he divides into classes all 
the inequalities of that planet, as far as they depend either on the elonga- 
tion of the sun and moon, or upon the eccentricity, or the parallax, or 
the inclination of the lunar orbit. The inherent difficulties of this task 
were immensely enhanced by the fact that Euler was virtually blind, and 
had to carry all the elaborate computations involved in his memory. A 
further difficulty arose from the burning of his house and the destruction 
of a greater part of his property in 1771. His manuscripts were fortu- 
nately preserved. His own life only was saved by the courage of a 
native of Basel, Peter Grimmon, who carried him out of the burning 
house. 

Some time after this, the celebrated Wenzell, by couching the cataract, 
restored his sight ; but a too harsh use of the recovered faculty, together 
with some carelessness on the part of the surgeons, brdught about a 
relapse. With the assistance of his sons, and of Krafft and Lexell, how- 
ever, he continued his labors, neither the loss of his sight nor the in- 
firmities of an advanced age being sufficient to check his activity. .Hav- 
ing engaged to furnish the Academy of St. Petersburg with as many 
memoirs as would be sufficient to complete its acts for twenty years after 
his death, he in seven years transmitted to the Academy above seventy 
memoirs, and left above two hundred more, which were revised and com- 
pleted by another hand. 

Euler's knowledge was more general than might have been expected 
in one who had pursued with such unremitting ardor, mathematics and 
astronomy, as his favorite studies. He had made considerable progress 
in medicine, botany, and chemistry, and he was an excellent classical 
scholar and extensively read in general literature. He could repeat the 
Aenied of Virgil from the beginning to the end without hesitation, and 
indicate the first and last line of every page of the edition which he used. 
But such lines from Virgil as, "The anchor drops, the rushing keel is 
staid," always suggested to him a problem and he could not help en- 
quiring what would be the ship's motion in such a case. 

Euler's constitution was uncommonly vigorous and his general health 
was always good. He was enabled to continue his labors to the very close 
of his life so that it was said of him, that he ceased to calculate and to 
breathe at nearly the same moment. His last subject of investigation 
was the motions of balloons, and the last subject on which he conversed 
was the newly discovered planet Herschel. 

On the 18th of September, 1783, while he was amusing himself at tea 
with one of his grandchildren, he was struck with apoplexy, which ter- 
minated the illustrious career of this wonderful genius, at the age of 
seventy-six. His works, if printed in their completeness, would occupy 
from 60 to 80 quarto volumes. However, no complete edition of Euler's 
writings has been published, though the work has been begun twice. 

He was simple,, upright, affectionate, and had a strong religious faith. 
His single and unselfish devotion to the truth and his joy at the dis- 
coveries of science whether made by himself or others, were striking attri- 
butes of his character. He was twice married, his second wife being a 
half-sister of his first, and he had a numerous family, several of whom 
attained to distinction. His eloge was written for the French Academy 
by Condorcet, and an account of his life, with a list of his works, was 
written by Von Fuss, the secretary of the Imperial Academy of St. 
Petersburg. 



BIOGRAPHY. 



449 



As has been said, Euler wrote an immense number of works, chief 
of which are the following: Introductio in Analysin infinitorum, 1748, 
which was intended to serve as an introduction to pure analytical mathe- 
matics. This work produced a revolution in analytical mathematics, as 
the subject of which it treated had hitherto never been presented in so 
general and systematic a manner. The first part of the Analysis Infini- 
torum contains the bulk of the matter which is to be found in modern 
text-books on algebra, theory of equations, and trigonometry. In the 
algebra, he paid particular attention to the expansion of various functions 
in series, and to the summation of given series, and pointed out explicitly 
that an infinite series can not be safely employed in mathematical inves- 
tigations unless it is convergent. In trigonometry, he introduced (simul- 
taneously with Thomas Simpson in England) the now current abbrevia- 
tions for trigonometric functions, and simplified formulas by the simple 
expedient of designating the angles of a triangle by A, B, C, and the op- 
posite sides by a, b, c. He also showed that the trigonometrical and ex- 
ponential functions are connected by the relation cos#-Hsin#=^'#. Here 
too we meet the symbol e used to denote the base of the Naperian 
logarithms, namely the incommensurable number 2.7182818 . . . and 
the symbol ?r used to denote the incommensurable number 3.14159265 . 
. . The use of a single symbol to denote the number 2.7182818 . . . 
seems to be due to Cotes, who denoted it by M. Newton was probably 
the first to employ the literal exponential notation, and Euler using the 
form a%, had taken a as the base of any system of logarithms. It is 
probable that the choice of e for a particular base was determined by 
its being the vowel consecutive to a, or, still more probable because e 
is the initial of the word exponent. 

The use of a single symbol to denote 3.14159265 . . . appears to 
have been introduced by John Bournilli, who represented it by c. Euler 
in 1734 denoted it by p, and in a letter of 1736 in which he enunciated the 
theorem that the sum of the square of the reciprocals of the natural 
numbers is ^-Tr 2 , he uses the letter c. The symbol TT was first used 
to represent 3.141592 ... by William Jones's in his "Synopsis Pal- 
mariorum Matheseos", London, 1706, and after the publication of Euler's 
Analysis, the symbol TT was generally employed, the choice of TT being 
determined by the initial of the word, Ttpi<p peta = periphereia. 

The second part of the Analysis Infinitorum is on analytical geometry. 
Euler begins this part by dividing curves into algebraic and transcendental, 
and establishes a number of propositions which are true for all algebraic 
curves. He then applied these to the general equation of the second 
degree in two dimensions, showed that it represents the various conic 
sections, and deduces most of their properties from the general 'equation. 
He also considered the classification of cubic, quartic, and other alge- 
braic curves. He next discussed the question as to what surfaces are 
represented by the general equation of the second degree in three dimen- 
sions, and how they may be discriminated one from the other. Some 
of these surfaces had not been previously investigated. In this work he 
also laid down the rules for the transformation of coordinates in space. 
Here also we find the first attempt to bring the curvature of surfaces 
within the domain of mathematics, and the first complete discussion of 
tortuous curves. 

In 1755 appeared Institutions Calculi Differential, to ^which the 
Analysis Infinitorum was intended as an introduction. This is the first 
text-book on the differential calculus which has any claim to be regarded 
as complete, and it may be said that most modern treatises on the sub- 
ject are based upon it. 

At the same time, the exposition of the principles of the subject is 
often prolix and obscure, and sometimes not quite accurate. 



450 FINKEVS SOLUTION BOOK. 

This series of works was completed by the publication in three vol- 
umes in 1768 to 1770 of the Institutiones Calculi Integralis, in which the 
results of several of Euler's earlier memoirs on the same subjects and 
on differential equations are included. In this treatise as in the one on. 
the differential calculus was summed up all that was at that time known 
on the subject. The beta and gamma functions were invented by Euler, 
and are discussed here, but only as methods of reduction and integration. 
His treatment of elliptic integrals is superficial. The classic problems on. 
isoperimetrical curves, the brachistochrone in a resisting medium, and 
theory of geodesies had engaged Euler's attention at an early date, and 
the solving of which led him to the calculus of variations. The general 
idea of this was laid down in his Curvarum Maximi Minimise Proprietate 
Gaudentium Inventio Nova ac Facilis, published in 1744, but the com- 
plete development of the new calculus was first effected by Lagrange in 
1759. The method used by Lagrange is described in Euler's integral 
calculus, and is the same as that given in most modern text-books on 
the subject. 

In 1770, Euler published the Anleitung zur Algebra in two volumes. 
The first volume treats of determinate algebra. This work includes the 
proof of the binomial theorem for any index, which is still known by 
Euler's name. The proof, which is not accurate according to the 
modern views of infinite series, depends upon the principle of the per- 
manence of equivalent forms, and may be seen in C. Smith's Treatise on 
Algebra, pages 336-7. Euler's proof with important additions due to 
Cauchy, may be seen in G. Chrystal's Algebra, Part II. 

It is a fact worthy of note that Euler made no attempt to investigate 
the convergency of the series, though he clearly recognized the necessity of 
considering the convergency of infinite series. While Euler recognized the 
convergency of series, his conclusions in reference to infinite series are not. 
always sound. In his time no clear notion as to what constitutes a con- 
vergent series existed, and the rigid treatment to which infinite series are 
now subjected was undreamed of. Euler concluded that the sum of the 
oscillating series 1 1 + 1 14-1 1+ . . . = , for the reason, that 
by stopping with an even number of terms the sum is 0, and by stopping 
with an odd number of terms the sum is 1. Hence, the sum of the series 
is (0 + 1) = . Guido Grandi went so far as to conclude that = + 
+ + . . . The paper in which Euler cautions against divergent 

series contains the proof that . . . y + 1- 1 + n + 2 + n 3 . . . = 0. 

His proof is as follows, n + n 2 + 3 + . . . 1 -I 1 h * 

1 n n n* 

= :-, : |- ~ = 0. Euler had no hesitation in writing 1 3 

"ft JL % J. JL 72 

_j_5_7 + 9 ... = 0, and he confidently believed that sin^ 2sin2^ + 
3sin3y? . . . = 0. 

A remarkable development, due to Euler, is what he named the hyper- 
geometrical series, the summation of which he observed to be dependent 
upon the integration of linear differential equations of the second order, 
but it remained for Gauss to point out that for special values of the 
letters, this series represented nearly all the functions then known. By 
giving the factors 641 X 6700417 of the number 2^ + 1 4294967297 
when n = 5, he pointed out the fact that this expression did not always 
represent primes, as was supposed by Fermat. By B. F. Finkel. From, 
the American Mathematical Monthly, Vol. IV, No. 12. 



BIOGRAPHY. 



451 



SOPHUS LIE.* 

Sophus Lie was born on the 17th of December, 1842, at Nordfjordeid 
(near Floro) where his father, John Herman Lie, was pastor. The 
studies of his childhood and youth did not reveal in him that exceptional 
aptitude for mathematics which is signalized so early in the lives of the 
great geometers: Gauss, Abel, and many others. Even on leaving the 
University of Christiania in 1865, he still hesitated between philology and 
mathematics. It was the works of Pliicker on modern geometry which 
first made him fully conscious of his mathematical abilities and awak- 
ened within him an ardent desire to consecrate himself to mathematical 
research. Surmounting all difficulties and working with indomitable 
energy he published his first work in 1869, and we can say that from 
1870 on he was in possession of the ideas which were to direct his 
whole career. 

At this time I frequently had the pleasure of meeting and conversing 
with him in Paris where he had come with his friend F. Klein. A course 
of lectures by Sylow revealed to Lie all the importance of the theory of 
substitution groups ; the two friends studied this theory in the great 
treatise of our colleague Jordan ; they saw fully the essential role which 
it would be called upon to play in all the branches of mathematics to 
which it had then not been applied. They have both had the good 
fortune to contribute by their works to impressing upon mathematical 
studies the direction which appeared to them to be the best. 

A short note of Lie "Sur une transformation geometrique," pre- 
sented to our Academy in October, 1870, contains an extremely original 
discovery. Nothing resembles a sphere less than a straight line and 
yet, by using the ideas of Pliicker, Lie found a singular transformation 
which makes a sphere correspond to a straight line, and which conse- 
quently makes possible the derivation of a theorem relative to an en- 
semble of spheres from every theorem relative to an aggregate of straight 
lines, and vice versa. It is true that if the lines are real, the correspond- 
ing spheres are imaginary. But such difficulties are not sufficient to 
deter geometers. In this curious method of transformation, each property 
relative to asymptotic lines of a surface is transformed into a property 
relative to lines of curvature. The name of Lie will remain attached 
to these concealed relations which connect the two essential and funda- 
mental elements of geometric investigation, the straight line and sphere. 
He has developed them in detail in a memoir full of new ideas which 
appeared in 1872 in the Mathematische. Annalen. 

The works following this brilliant beginning fully confirmed all the 
hopes to which it gave birth. Since the year 1872 Lie has put forth a 
series of memoirs upon the most difficult and most advanced parts of the 
integral calculus. He commences by a profound study of the works of 
Jacobi on the partial differential equations of the first order and at first 
cooperates with Mayer in perfecting this theory in an essential point. 
Then, by continuing the study of this beautiful subject, he is led to 
construct progressively that masterful theory of continuous transforma- 
tion groups which constitutes his most important work and in which, at 
least at the start, he was aided by no one. The detailed analysis of 
this vast theory would require too much space here. It is proper, how- 
ever, to point out particularly two elements wholly essential to these 
researches: first, the use of contact transformations which throws such 

-'From the Bulletin of the American Mathematical Society. Translated by Edgar 
Odell I<ovett from Comptcs Rendus. 



452 FINKBIv'S SOLUTION BOOK. 

a vivid and unexpected light upon the most difficult and obscure parts 
of the theories relative to the integration of partial differential equations; 
second, the use of infinitesimal transformations. The introduction of 
these transformations is due entirely to Lie; their use, like that of 
Lagrange's variation, naturally greatly extends both the notion of differ- 
ential and the applications of the infinitesimal calculus. 

The construction of so extended a theory did not satisfy Lie's ac- 
tivity. In order to show its importance he has applied it to a great num- 
ber of particular subjects, and each time he has had the good fortune of 
meeting with new and elegant properties. I find my preference in the 
researches which he has published since 1876 on minimal surfaces. The 
theory of these surfaces, the most attractive perhaps that presents itself 
in geometry, still awaits, and may await a long time, the complete solu- 
tion of the first problem to be proposed in it, namely, the determina- 
tion of a minimal surface passing through a given contour. But, in re- 
turn, it has been enriched by a great number of interesting propositions 
due to a multitude of geometers. In 1866 Weierstrass made known a 
very precise and simple system of formulae which has called forth a 
whole series of new studies on these surfaces. In his works Lie returns 
simply to the formulae of Monge ; he gives their geometric interpreta- 
tion and shows how their use can lead to the most satisfactory theory 
of minimal surfaces. He makes known methods which permit of deter- 
mining all algebraic minimal surfaces of given class and order. Finally, 
he studies the following problem: to determine all algebraic minimal 
surfaces inscribed in a given algebraical developable surface. He gives 
the complete solution for the case where only one of these surfaces in- 
scribed in the developable is known. 

Of great interest also are the researches which we owe to him on 
the surfaces of constant curvature, in the study of which he makes use 
of a theorem of Bianchi on geodesic lines and circles, likewise those on 
surfaces of translation, on the surfaces of Weingarten, on the equations 
of the second order having two independent variables, et cetera. I should 
reproach myself for forgetting, even in so rapid a resume, the appli- 
cations which Lie has made of his theory of groups to the non-Euclidean 
geometry and to the profound study of the axioms which lie at the basis 
of our geometric knowledge. 

These extensive works quickly attracted to the great geometer the 
attention of all those who cultivate science or are interested in its prog- 
ress. In 1877 a new chair of mathematics was created for him at the 
University of Christiania, and the foundation of a Norwegian review 
enabled him to pursue his work and publish it in full. In 1886 he 
accepted the honor of a call to the University of Leipzig ; he taught in 
this university with the rank of ordinary professor from 1886 to 1898. 
To this period of his life is to be referred the publication of his didactic 
works, in which he has coordinated all his researches. Six months ago 
he returned to his native land to assume at Christiania the chair which 
had been especially reserved for him by the Norwegian parliament, with 
the exceptional salary of ten thousand crowns. Unfortunately, excess 
of work had exhausted his strength and he died of cerebal anaemia at the 
age of fifty-six years. 

Nowhere is his loss felt more keenly than in our country, where he 
had so many friends. True, in 1870 a misadventure befell him, whose 
consequences I was instrumental in averting. Surprised at Paris by the 
declaration of war, he took refuge at' Fontainebleau. Occupied in- 
cessantly by the ideas fermenting in his brain, he would go every day 
into the forest, loitering in places most remote from the beaten path, 
taking notes and drawing figures. It took little at this time to awaken 
suspicion. Arrested and imprisoned at Fontainebleau, under conditions 
otherwise very comfortable, he called for the aid of Chasles, Bertrand, 
and others; I made the trip to Fontainebleau and had no trouble in 



BIOGRAPHY. 453 

convincing the procureur imperial; all the notes which had been seized 
and in which figured complexes, orthogonal systems, and names of ge- 
ometers, bore in no way upon the national defenses. Lie was released; 
his high and generous spirit bore no grudge against our country. Not 
only did he return voluntarily to visit it but he received with great 
kindness French students, scholars of our Ecole Normale who would 
go to Leipzig to follow his lectures. It is to the Ecole Normale that 
he dedicated his great work on the theory of transformation groups. A 
number of our thesis at the Sorbonne have been inspired by his teaching 
and dedicated to him. 

The admirable works of Sophus Lie enjoy the distinction, to-day 
quite rare, of commanding the common admiration of geometers as well 
as analysts. He has discovered fundamental propositions which will pre- 
serve his name from oblivion, he has created methods and theories which, 
for a long time to come, will exercise their fruitful influence on the 
development of mathematics. The land where he was born and which 
has known how to honor him can place with pride the name of Lie be- 
side that of Abel, of whom he was a worthy rival and whose approaching 
centenary he would have been so happy in celebrating. By Professor 
Gaston Darboux. 



454 FINKEI/S SOLUTION BOOK. 



BIOGRAPHY. 



SIMON NEWCOMB, PH. D M LL. D. 

Simon Newcomb was born in Wallace, Nova Scotia, in 1835. After 
being educated by his father he engaged for some time in teaching. He 
carne to the United States in 1853, and was engaged for two years as 
a teacher in Maryland. There he became acquainted with Joseph Henry 
and Julius E. Hilgard, who recognizing his aptitude for mathematics, 
secured his appointment in 1857 as computer on the "Nautical Almanac," 
which was then published in Cambridge, Mass. In Cambridge he came 
under the influence of Professor Benjamin Peirce. He entered the Law- 
rence Scientific School and was graduated in 1858, continuing thereafter 
for three years as a graduate student. 

In 1861 he was appointed professor of mathematics in the U. S. 
Navy and assigned to duty at the U. S. Naval Observatory in Wash- 
ington. There he negotiated the contract for the 26-inch equatorial tele- 
scope authorized by congress, supervised its construction and planned the 
tower and dome in which it is mounted. 

He was chief director of the commission created by congress to ob- 
serve the transit of Venus on December 8, 1874. He visited the Sas- 
katchewan region in 1860 to observe an eclipse of the Sun, and in 1870-1 
was sent to Gibralter for a similar purpose. In 1882 he commanded an 
expedition to observe the transit of Venus at the Cape of Good Hope. 
Meanwhile in 1887 he became senior professor of mathematics in the U. S. 
Navy, and since that time has been in charge of the office of the "Amer- 
ican Ephemeris and Nautical Almanac." Professor Newcomb has a large 
corps of assistants in Washington. 

In addition to these duties, in 1884 he became professor of mathe- 
matics and astronomy in Johns Hopkins, (succeeding the distinguished 
Sylvester, upon the departure of the latter to accept a professorship at 
Oxford), where he has had charge of the American Journal of Mathe- 
matics. However he is not now editor of that Journal, having recently 
severed his immediate active connection with the Johns Hopkins Uni- 
versity for the next two or three years. 

Professor Newcomb has been intimately associated with the equip- 
ment of the Lick observatory of California, and examined the glass of 
the great telescope and its mounting before its acceptance by- the trustees. 

The results of his scientific work have been given to the world in 
more than one hundred papers and memoirs. Concerning these, Arthur 
Cayley, president of the Royal Astronomical Society of Great Britain, said: 
"Professor Newcomb's writings exhibit, all of them, a combination on 
the one hand of mathematical skill and power and on the other of good 
hard work, devoted to the furtherance of astronomical science." 

His work has been principally in the mathematical astronomy of the 
solar system, particularly Neptune, Uranus, and the Moon, but the whole 
plan includes the most exact possible tables of the motions of all the 
planets. Amongst the most important of his papers are : "On the Secu- 
lar Variations and Mutual Relations of the Orbits of the Asteroids" 
(1860) ; "An Investigation of the Orbit of Neptune, with general tables 
of its motion" (1874) ; "Researches on the Motion of the Moon" (1876) ; 
"Measure of the Velocity of Light" (1884) ; and "Development of the 
Purturbative Function and its Derivative in the Sines and Cosines of the 
Eccentric Anomaly, and in Powers of the Eccentricities and Inclinations" 
(1884). 

In 1874 Columbian University of Washington conferred on him 'the 
degree of LL. D., and in 1875 he received the same degree from Yale,, 



BIOGRAPHY. 455 

also from Harvard in 1874, and from Columbia College in 1887, while on 
the 300th anniversary of the founding of the University of Leyden in 
1875, that institution gave him him the degree of Master of Mathematics 
and Doctor of Natural Philosophy, and on the 500th anniversary of the 
University of Heidelberg in 1886 he received the degree of Ph. D. Be- 
sides the degrees just mentioned he received one from Edinburgh in 1891, 
one on the occasion of the tercentenary of the University of Dublin in 
1892, and one from Paris on the tercentenary of Galileo's connection with 
the University in 1893. 

He was awarded the gold medal of the Royal Astronomical Society in 
1874 and in 1878 received the great gold Huyghens medal of the Uni- 
versity of Leyden, which is given to astronomers once in 20 years for 
the most important work accomplished in that science between its 
awards. Besides the two gold medals mentioned Professor Newcomb 
received a third in 1890, the Copley medal, given by the Royal Society 
of England. 

In 1887 the Russian Government ordered the portrait of Professor 
Newcomb to be painted for the collection of famous astronomers at the 
Russian observatory at Pulkowa, and also ordered to be presented to 
him a vase of jasper with marble pedestal seven feet high. The Univer- 
sity of Tokyo has also presented him with two vases of bronze. 

He was elected an associate member of the Royal Astronomical So- 
ciety in 1872, corresponding member of the Institute of France in 1874, 
and foreign member of the Royal Society 1877 ; and he also holds hon- 
orary or corresponding relations to nearly all the European academies of 
Science. In 1877 he was elected one of the eight members of the council 
of the Astronomische Gesellschaft, an international astronomical society 
that meets once in two years. He was elected to the National Academy 
of Sciences in 1869 and since 1883 has been its vice president. In 1876 
he was elected president of the American Association for the Advance- 
ment of Science, and delivered his retiring address at the St. Louis 
meeting in 1878. He also held the presidency of the American Society 
for Physical Research. 

He was elected member of the New York Mathematical Society in 
1891, and delivered an address, entitled "Modern Mathematical Thought" 
before the annual meeting of the Society, December 28, 1893, which was 
published in the Bulletin of the Society for January 1894, and in Nature 
of February 1, 1894. 

Professor Newcomb's book on Popular Astronomy (1877) has been 
republished in England and translated into German, while "School As- 
tronomy" by Newcomb and Holden (1879), and their "Briefer Course" 
(1883), are used as text books in most of our colleges. 

Professor Newcomb has also carried on important investigations on 
subjects purely mathematical. An important contribution by him on 
"Elementary Theorems Relating to the. Geometry of a Space of Three 
Dimensions and of Uniform Positive Curvature in the Fourth Dimen- 
sion," was published in Borchardt's Journal, Berlin, 1877. Full extracts 
of this important contribution to non-Euclidean geometry are given in 
the Encyclopaedia Britannica, article "Measurment." In Vol. I. of the 
American Journal of Mathematics he has a note "On a Class of Trans- 
formations which Surfaces may Undergo in Space of more than Three 
Dimensions," in which he shows, for instance, that if a fourth dimen- 
sion were added to space, a closed material surface (or shell) could be 
turned inside out by simple flexure without either stretching or tearing. 
Later articles have been on the theory of errors in observations. In 
former years he also contributed to the Mathematical Monthly and the 
Analyst. 

He has also written a series of mathematical text-books, comprising 
Algebra (1881) ; Geometry (1881) ; Trigonometry and Logarithms 



456 FINKEI/S SOLUTION BOOK. 

(1882) ; School Algebra (1882) ; Analytic Geometry (1884) ; Essentials 
of Trigonometry (1884); and Calculus (1887). These works have been 
favorably received and are everywhere regarded as text-books of decided 
merits. 

Professor Newcomb refers to astronomy as his profession and to polit- 
ical economy as his recreation, and in the latter branch has written several 
books and a number of magazine articles. By J. M. Colaw. From the 
American Mathematical Monthly, Vol. I, No. 8. 



BIOGRAPHY. 



457 



GEORGE BRUCE HALSTED. 

Dr. Halsted, a direct descendant of Abram Clark, signer of the 
declaration of independence, was born in Newark, N. J., November 25, 
1853. The Halsteds have been for four generations graduates of Prince- 
ton. Winning the Mathematical Fellowship at Princeton, Halsted went 
for applied mathematics to Columbia School of Mines, won while there 
an intercollegiate prize, and was made one of the first Fellows of the 
new Johns Hopkins University. Sylvester was particularly partial to 
him, and Halsted's life of Sylvester is recognized as the authority on 
the American part of his career. Says Dr. Fabian Franklin: "Professor 
Halsted, in his account of Sylvester's work already referred to, points out 
how the vicissitudes of his career were reflected in the richness or the 
meagreness of his mathematical production from period to period." 
Says Major P. A. MacMahon in the biography of Sylvester published 
by the Royal Society: "Sylvester's first high class consisted of but one 
student, G. B. Halsted. This gentleman, since well known in science, 
had the most beneficial effect upon his master, for it was owing to his 
enthusiasm and persistence that Sylvester's attention was again called 
to the Modern Higher Algebra and the Theory of Invariants, and a fruit- 
ful crop of new discoveries was almost the immediate result. Before 
taking his Doctor's Degree at the Johns Hopkins University, Halsted 
published in 1878 his epoch-making Bibliography of Hyper-Space and 
Non-Euclidean Geometry, long out of print and greatly in demand, which 
began at once to be cited all over the world and was reproduced in 
Russia. Called to introduce modern higher mathematics at his ancestral 
college, Princeton, the papers set on quaternions, determinants, modern 
higher algebra, and history of mathematics by Dr. Halsted were the 
first ever given at Princeton. His teaching of the History of Mathe- 
matics attracted attention as early as 1881, when he had in his class Pro- 
fessor H. B. Fine and Professor A. L. Kimball. His work in history of 
mathematics has never since been discontinued. While at Princeton, Dr. 
Halsted produced his treatise on Mensuration which had the honor of 
being drawn upon by Professor Wm. Thomson for his article "Mensura- 
tion" in the ninth edition of the Encyclopaedia Britannica. He took 
from it, among other novelties, the steregon, the steradian, and the treat- 
ment of solid angles associated with them. Simon Newcomb wrote of 
Dr. Halsted: "He is the author of a treatise on Mensuration which is 
the most thorough and scientific with which I am acquainted." 

From this brilliant field of work Dr. Halsted was called by a tele- 
gram from the University of Texas, announcing his election to the pro- 
fessorship of mathematics, urging his acceptance. Dr. Halsted had al- 
ready produced 17 scientific papers on Mathematics and Logic. This 
productivity was not abated by his removal to Texas, and the titles of his 
papers now approach a hundred. 

His Elements of Geometry which appeared in 1885, has passed 
through seven editions. The section headed by his phrase "Partition of 
a Perigon," has become classic. 

Of Dr. Halsted's Elementary Synthetic Geometry, which appeared 
in 1893, has been said: "For more than two thousand years geometry 
has been founded upon, and built up by means of, congruent triangles. 
At last, after twenty centuries, comes a book reaching all the preceding 
results without making any use of congruent triangles; and so simply 
that, for example, all the ordinary cases of congruence of triangles are 
demonstrated together in eight lines." 



458 FINKEIv'S SOLUTION BOOK. 

Dr. Halsted's two books are as yet the only geometries treating 
spherics comparatively by considering the sphere as the analogue of 
the plane. "In the method employed by Dr. Halsted almost the whole of 
the geometry of the plane is directly applicable to the sphere." The Ele- 
mentary Synthetic Geometry contains an introduction to the new Le- 
moine-Brocard Geometry, the only one which has appeared on the 
western continent. 

David Eugene Smith, in his history of Modern Mathematics men- 
tions, p. 567, as "among the most -active" in securing the acceptance of 
the Bolyai-LobachevskHdea, Houel in France, Riemann, Helmholtz and 
Baltzer in Germany, de Tilley (1879) in Belgium, Clifford in England, 
and Halsted (1878) in America. 

Of these all are dead but Halsted. By Halsted alone were the im- 
mortal works of Saccheri, Bolyai, Lobachevski, Vasiliev, made available 
to the English speaking world, and also Japan, for his Bolyai and Lo- 
bachevski have been reissued in Tokio. Results of his travels in Japan, 
Hungary, Russia, added fuel to the awakening fire of interest in all things 
non-Euclidean, and now the majestic face of heroic Lobachevski has 
t>een worthily given to the world in the beautiful picture made from a 
photograph furnished by Dr. Halsted to the Open Court as frontispiece 
for his Life of Lobachevski in the number for July 1898. 

Of John Bolyai he could find in all Hungary no picture; his linea- 
ments are lost forever. 



BIOGRAPHY. 
BIOGRAPHY, 



PROFESSOR FELIX KLEIN. 

The eminent subject of this very imperfect sketch was born on the 
.twenty-fifth of April, 1849, in Duesseldorf. His mother was Elise Sophie 
nee Kayser; his father, the "Landrentmeister" Caspar Klein, both of 
the protestant faith. For eight years, from the autumn of 1857 to the 
autumn of 1865 he attended the Duesseldorf Gymnasium, and went thence 
to the University of Bonn, for the study of mathematics and the natural 
sciences, especially physics. Here he had the extraordinary good fortune 
to come into close relations with the great Professor Pluecker, who gave 
him the position of assistant in the physical institute of Bonn, and used 
his help in writing out his profoundly original and stimulating mathe- 
matical works. 

The death of Pluecker May 22nd, 1868, closed this formative period, 
of which the influence on Klein can not be overestimated. So mighty is 
the power of contact with the living spirit of research, of taking part in 
original work with a master, of sharing in creative authorship, that any 
one who has once come intimately in contact with a producer of the 
first rank must have had his whole mentality altered for the rest of 
his life. 

The gradual development, high attainment, and then continuous 
achievement of Felix Klein are more due to Pluecker than to all other 
influences combined. His very mental attitude in the world of mathe- 
matics constantly recalls his great maker. 

Of others whose lectures he attended, we may mention Argelander 
<and Lipschitz, to the latter of whom particularly he has expressed his 
gratitude for kindly and efficient guidance and aid in his studies. Klein 
took his doctor's degree at Bonn on December 12th, 1868, with a disser- 
tation "On the transformation of the general equation of the second 
degree between line-coordinates to a canonic form," a subject taken from 
the analytic line-geometry of his master Pluecker. A line-complex of the 
nth degree contains a triply infinite multitude of straights, which are 
so distributed in space, that those straights which go through a fixed 
point make a cone of the nth order, or, what is the same, that those 
straights which lie in a fixed plane envelop a curve of the nth class. Such 
an aggregate or form finds its analytic representation through the coordi- 
nates of the straight in space, introduced by Pluecker. According to 
Pluecker the straight has six homogeneous coordinates which fulfill an 
equation-pf-conditipn of the second degree. By means of these the 
straight is determined with reference to a coordinate-tetrahedron. A 
homogeneous equation of the nth degree between these coordinates rep- 
resents a complex of the nth degree. 

The dissertation transforms the equation of the second degree be- 
tween line-coordinates to a canonic form, in correspondence with a 
change of the coordinate-tetrahedron. It first gives the general formulas 
to be applied in such a transformation. 

From these the problem appears algebraically as the simultaneous 
linear transformation of the complex to a canonic form, and of theequation- 
'pf-condition, which the line coordinates must fulfill, into itself. In carry- 
ing out these transformations, it attains to a classification of the com- 
plexes of the second degree into distinct species. 

The dissertation is dedicated to Pluecker and contains eight specific 
references to Pluecker's "Neue Geometric des Raumes, gegruendet auf die 
Betrachtung der geraden Linie als Raumelement." It is lucid and simple, 
but for depth and promise contrasts sharply with the great dissertation 
of Riemann, that "book with seven seals." 



460 FINKEI/S SOLUTION BOOK. 

It may be interesting, as characteristic of this germinating state, to 
note that of his five theses the second calls attention to one of Cauchy's 
slips in logical rigor, slips now known to be so numerous that C. S. Pierce 
makes of them a paradox, maintaining that fruitfulness of Cauchy's work 
is essentially connected with its logical inaccuracy. 

The third thesis declares the assumption of an ether unavoidable in 
the explanation of the phenomena of light. 

The last thesis is the desirability of the introduction of newer methods 
in Geometry alongside the Euclidean in gymnasial teaching. 

This serves, it seems, to emphasize my point that the long eight years 
of gymnasial so-called training left the seed still dormant, and only in 
Pluecker did it find the rain and the sun to call it to life and growth. 

Within two years now the development is amazing. Already in 1870 
he is working with another great genius, Sophus Lie ; and in 1871 is 
presented to the Goettingen Academy of Science his epoch-making paper, 
"Ueber die sogenannte Nicht-Euklidische Geometric." Its aim is to pre- 
sent the mathematical results of the non-Euclidean geometry, in so far 
as they pertain to the theory of parallels, in a new, intuitive way ; its in- 
strument is the mighty projective geometry, which he proves independent 
of all question of parallels. He perfects the projective metrics of Cay- 
ley by founding cross- ratio, after von Staudt, wholly without any use or 
idea of measurement. Then can be constructed a general projective ex- 
pression for distance, related to an arbitrary surface of the second degree 
as Fundamental-surface (Cayley's Absolute). This projective metrics 
then gives, according to the species of Absolute used, a picture of the 
results of the parallel-theory in the space of Lobachevsky, of Euclid, of 
Riemann. But not merely a picture ; they coincide to their innermost 
nature. 

The paper begins by stating that, as well-known, the eleventh axiom 
of Euclid is equivalent to the theorem that the sum of the angles in a 
triangle equals two right angles. Legendre gave a proof that the angle- 
sum in a triangle cannot be greater than two right angles ; but this proof, 
like the corresponding one in Lobachevsky, assumes the infinite length of 
the straight. 

Drop this assumption, and the proof falls, else would it apply in 
surface spherics. Legendre showed further, that if in one triangle the 
angle-sum is two right angles, it is so in every triangle. We now know 
that this had been proven long before by Saccheri. But Professor Klein 
said that he heard the name of Saccheri for the first time in my address 
before the World's Science Congress. But it is claimed for Gauss that he 
was the first to distinctly state his conviction of the impossibility of prov- 
ing the theorem of the equality of the angle-sum to two right" angles. 
But it does not follow, as claimed by his Goettingen worshippers, that 
Gauss ever came to the ^ conviction that a valid non-Euclidean geometry 
was possible until after it had been made simultaneously by John Bolyai 
and Lobachevsky, and perhaps long before by Wolfgang Bolyai. Cer- 
tainly the world did not hear of it from Gauss. He published nothing 
on it. 

In this non-Euclidean geometry there appears a certain constant 
characteristic for the metrics of the space. By giving this an infinite 
value we obtain the ordinary Euclidean geometry. But if it has a finite 
value, we get a quite distinct geometry, in which, for example, the follow- 
ing theorems hold: The angle-sum in a triangle is less than two right 
angles, and indeed so much the more so the greater the surface of the 
triangle. For a triangle whose vertices are infinitelv separated, the angle- 
sum is zero. Through a point without a straight one can draw two 
parallels to the straight, that is, lines which cut the straight on the one 
or the other side in a point at infinity. The straights through the point 
which run between the two parallels nowhere cut the given straight. But 
on the other hand, in Riemann' s marvellous inaugural lecture, "Ueber die- 



BIOGRAPHY. 



461 



Hypothesen, welche der Geometric zu Grunde liegen," is pointed out that 
the unboundedness of space, which is experiential, does not carry with it 
the infinity of space. 

It is thinkable, and would not contradict our perceptional intuition, 
which always relates to a finite piece of space, that space is finite and 
comes back into itself. 

The geometry of our space would then be like that of a tridimensioual 
sphere in a four dimensional manifoldness. This representation carries 
with it that the angle-sum in a triangle, as in ordinary spherical triangles, 
is greater than two right angles, and indeed the more so, the greater the 
triangle. The straight would then have no point at infinity, and through 
a given point no parallel to a given straight could be drawn. Now Cayley 
constructed his celebrated projective metrics to show how the ordinary 
Euclidean metrics may be taken as a special part of projective geometry. 
Klein generalizes Cayley and founds three metric geometries, the elliptic 
(Riemann's), the hyperbolic (Lobachevsky's), the parabolic (Euclid's). 

This little paper of 1871 contains the promise of much that is most 
genial in the after work of a man now generally considered as the most 
interesting and one of the very greatest of living mathematicians. Of all 
those splendid and charming series of lectures with which Klein has made 
Goettingen so attractive to the whole world, the most delightful and 
epoch-making are those on non-Euclidean geometry, (Nicht-Euklidische 
Geometric, I. Vorlesung, gehalten waehrend des Wintersemesters 1889-90 
von F. Klein. Ausgearbeitet von Fr. Schilling. Zweiter Abdruck. Goet- 
tingen, 1893. Small Quarto, lithographed, pp. v. 365. II. Sommersemes- 
ters 1890. Zweiter Abdruck 1893, pp. iv. 238). 

The World's Science Congress at Chicago was in nothing more fortu- 
nate than in the presence of Helmholtz and Felix Klein, and in the spon- 
taneous and universal homage accorded them no idea was more often 
emphasized than their connection with the birth and development of that 
wonderful new world of pure science typified in the non-Euclidean 
geometry. 

The narrow limits of this feeble sketch prevent the statement of how 
much promise, richly fulfilled in the development of this many-sided man, 
in totally other directions is contained in a little-known paper of 1873, 
"Ueber den allgemeinen Functionsbegriff und dessen Darstellung durch 
eine willkuerliche Curve." 

Twenty years of production and achievement have not in the least 
dampened the ardour of this enthusiastic mind. This very summer at the 
great meeting of scientists in Vienna Klein seemed the busiest, the fore- 
most of all that goodly company. By Dr. George Bruce Halsted. From 
the American Mathematical Monthly, Vol. I, No. 12. 



462 FINKEL'S SOLUTION BOOK. 

BIOGRAPHY. 



BENJAMIN PEIRCE. 

Benjamin Peirce was born at Salem, Massachusetts, April 4, 1809, 
and died at Cambridge, Massachusetts, October 6, 1880. He entered Har-. 
vard College, at the age of sixteen; and, at the age of twenty, he was 
graduated from the same College, with highest honors. He devoted him- 
self principally to the study of Mathematics. This favorite study of hia 
was pursued far beyond the limits of the curriculum of mathematical 
studies prescribed by the authorities of Harvard College, at that time. 

As an under-graduate student, young Peirce was instructed by 
Nathaniel Bowditch, who soon perceived the innate mathematical genius 
of his pupil. Bowditch proudly predicted the future greatness of the 
young man. Not only did Bowditch give him valuable instruction in 
geometry and analytics, but also acted as his mathematical adviser 
carefully directing him in the development of his mathematical talents and 
scientific powers. The lectures on higher mathematics delivered by Francis 
Grund he was enabled to attend, by reason of his preparation beyond the 
limit of the under-graduate course in mathematics. When Dr. Bowditch 
was publishing his translation and commentary of the Mehanique Celeste 
of Laplace, young Peirce assisted in reading the proof-sheets. This crit- 
ical reading of that great work of Laplace was to him an education in 
itself, and may have been the prime cause that not a small part of Peirce's 
subsequent mathematical and scientific work was done in the great field 
of analytical mechanics : 

In the class-room, he frequently gave original demonstrations which 
proved to be more direct and scientific than those given in the text-books 
of that day. On graduating, he went to Northampton, Massachusetts, as 
a teacher in Mr. Bancroft's School. As tutor, he returned to Harvard 
College, in 1831. Since Professor Farrar spent the next year in Europe, 
tutor Peirce was left at the head of the Department of Mathematics in 
Harvard College; and, on account of the physical inability of Professor 
Farrar to resume teaching, Peirce continued to fill his place. In fact, 
Peirce held this position, advancing step by step, until the time of his 
death. His position, in 1842, was christened "The Perkins Professorship 
of Mathematics and Astronomy." In the history of mathematical teach- 
ing at Harvard College, the year 1833 marks an important epoch; as it 
was then that Benjamin Peirce became the professor of Mathematics and 
Natural Philosophy in that institution of learning. 

Professor Peirce was married in July, 1833. At the time of his death, 
there were living his wife, three 'sons, and a daughter. His eldest son, 
James M. Peirce, is University professor of mathematics in Harvard; 
Charles S. Peirce is a professor in the Johns Hopkins University; and 
H. H. D. Peirce is connected with the firm of Herter Brothers, New 
York City. 

It has been said that a mere boy detected an error in Bowditch's solu- 
tion of a problem. "Bring me the boy who corrects my mathematics," 
said Bowditch. Master Benjamin Peirce was the boy who had done the 
correcting; and thirty years later, this same Benjamin Peirce dedicated 
one of his great mathematical works "To the cherished and revered 
memory of my master in science, Nathaniel Bowditch, The Father of 
American Geometry." This same title was bestowed upon Peirce, by for- 
eign mathematicians. Sir. "Win. Thomson (Lord Kelvin), in an address 
before the British Association, referred to Benjamin Peirce as "The 
Founder of High Mathematics in America;" and on a similar occasion, 
the late Professor Cayley referred to him as "The Father of American 
Mathematics." The name of Benjamin Peirce is that of an American 



BIOGRAPHY. 463 

mathematician, whom no one need hesitate to rank with the names of 
Pythagoras, Leibnitz, Newton, Legendre, John Bernoulli, Wallis, Abel, 
Laplace, Lagrange, and Euler. Through the united efforts of the late 
Professor Wm. Chauvenet (Yale's ablest mathematician and astronomer) 
and Benjamin Peirce not to speak of their worthy successors, was 
effected the general adoption of the ratio-system in American works on 
trigonometry. 

In the reforms incident to the New Education, Harvard has always 
taken a prominent part and Benjamin Peirce was an enthusiastic advocate 
of the elective system with respect to collegiate studies. As a branch of 
Harvard College, there was opened, in 1842, the Lawrence Scientific 
School ; and in this school, Professor Peirce gave instruction in higher 
Mathematics including analytical and celestial mechanics. Such advanced 
courses of mathematics, as he offered to students, in 1848, had never 
before been offered to American students by any other professor in any 
other American college. The second American educational institution 
which offered equally advanced courses of mathematics, is the Johns 
Hopkins University; and these courses were arranged by that English 
master, who gave a fresh and powerful impulse to mathematical study 
and teaching in America Professor J. J. Sylvester. 

The preparation of mathematical text-books was begun by Professor 
Peirce, immediately on beginning his career as teacher of Mathematics in 
Harvard College. In 1835 appeared his Elementary Treatise on Plane 
Trigonometry; in 1836, his Elementary Treatise on Spherical Trigo- 
nometry together with his Elementary Treatise on Sound; in 1837, his 
Elementary Treatise on Plane and Solid Geometry together with his Ele- 
mentary Treatise on Algebra; during the period of 1841^6, he wrote 
and published in two volumes his Elementary Treatise on Curves, Func- 
tions, and Forces; and in 1855, he published his Analytical Mechanics. 
Subsequently was published his memoir on Linear Associative Algebra; 
and this memoir, according to Professor James Mills Peirce, he regarded 
as his great work. All of his works are models of conciseness, perspicuity, 
and elegance; and they all evince extraordinary originality and genius. 

In 1867, Professor Peirce was made the Superintendent of the United 
States Coast Survey; and he held that position for seven years. He had 
been consulting astronomer to the American Ephemeris and Nautical Al- 
manac, since 1849; and for many years, he directed the theoretical part 
of the work. In 1855, Professor Peirce was one of the men intrusted 
with the organization of the Dudley Observatory. For many years before 
and after he took charge of the United States Coast Survey, he was 
frequently consulted with respect to the work in that office. He received 
the degree of Doctor of Laws from the University of North Carolina, in 
1847, and also from Harvard University in 1867. He was elected an As- 
sociate of the Royal Astronomical Society of London in 1849, and a 
member of the Royal Society of London in 1852. He was elected presi- 
dent of the American Association for the Advancement of Science, in 
1853 (the fifth year of its existence) ; and he was one of the original 
members of the Royal Societies of Edinburg, and Goettingen; Honorary 
Fellow of the Imperial University of St. Vladimir, at Kiev ; " etc. 

Professor Peirce's conception of the American Social Science Asso- 
ciation was that it should be a university for the people, combining those 
who can contribute any thing original in social science into a temporary 
academical senate, to meet for some weeks in a given place and debate 
questions with each other, as well as to give out information for the 
public. In this line of thought he favored, also, the establishment of the 
Concord School of Philosophy, to do a similar work in the speculative 
studies ; and he lived to see the partial realization of what he foresaw in 
this instance. In a Mathematical Society over which he presided forborne 
years, each member would bring something novel in his own particular 
branch of study; and in the discussion which followed, it would almost 



464 FINKEL'S SOLUTION BOOK. 

invariably appear that Professor Peirce had, while the paper was being^ 
read, pushed out the author's methods to far wider results than the 
author had dreamed possible. The same power of extending rapidly in his 
own mind novel mathematical researches was exhibited at the sessions of 
every scientific body at which he chanced to be present. What was quite 
as admirable was the way in which he did it, giving the credit of the 
thought always to the author of the essay under discussion. His pupils 
thus frequently received credit for what was in reality far beyond their 
attainment. He robbed himself of fame in two ways: by giving the 
credit of his discoveries to those who had merely suggested the line of 
thought and by neglecting to write out and publish that which he had 
himself thought out. 

In physical astronomy, perhaps, his greatest works were in connection 
with the planetary theory, his analysis of the Saturnian system, his re- 
searches regarding the lunar theory, and the profound criticism of the dis- 
covery of Neptune following the investigations of Adams and Leverrier. 
At the time of the publication of his "System of Analytical Mechanics," 
Professor Peirce announced that the volume would be followed by three 
others, entitled respectively: "Celestial Mechanics," "Potential Physics," 
and "Analytical Morphology." These three volumes were never pub- 
lished. 

Professor Peirce, in a paper read before the American Association 
for the Advancement of Science, in 1849, showed in the vegetable world 
the demonstrable presence of an intellectual plan showed that phyllotaxis 
(the science of the relative position of leaves) involved an algebraic idea; 
and this algebraic idea was subsequently shown to be the solution of a 
physical problem. 

The higher mathematical labors of so eminent a geometer must lie be- 
yond the course of general recognition. Among the things which give 
him a just claim to this title, may be mentioned: his discussion of the 
motions of two pendulums attached to a horizontal cord; of the motions 
of a top; of the fluidity and tides of Saturn's rings; of the forms of 
fluids enclosed in extensible sacs ; of the motions of a sling ; of the orbits 
of Uranus, Neptune, and the comet of 1843; of the criteria for rejecting 
doubtful observations; of a new form of binary arithmetic, of systems 
of linear and associative algebra ; of various mechanical games, puzzles, 
etc.; of various problems in geodesy; of the lunar tables; of the occul- 
tations of the Pleiades; etc. He adapted the epicycles of Hipparchus 
to the analytical_ forms of modern science; and he, also, solved by a 
system of co-ordinates of his own devising, several problems concerning 
the involutes and evolutes of curves, which would probably have proved 
impregnable by any other method of mathematical approach. 

None of Professor Peirce's labors lie farther above the ordinary reach 
of thought than his little lithographed volume on Linear and Associative 
Algebra. In this he discusses the nature of mathematical methods; and 
the characteristics which are necessary to give novelty and unity to a cal- 
culus. Then he passes to a description of seventy or eighty different 
kinds of simple calculus. Almost no comment is given; but the mathe- 
matical reader discovers, as he proceeds, that only three species of calcu- 
lus, having each a unity in itself, have been hitherto used to any great 
extent, namely, ordinary algebra, differentials, and quaternions. Think 
of it; what a wonderful volume of prophecy that is which describes sev- 
enty or eighty species of algebra, any one of which would require genera- 
tion after generation of ordinary mathematicians to develop! 

On both sides of the Atlantic, Professor Peirce as an author, was 
highly esteemed. His work on analytical mechanics was, at the time of its 
publication, regarded even in Germany, as the best of its kind. As a 
lecturer, Professor Peirce was highly esteemed in both scientific and popu- 
lar circles. It is related that in 1843, by a series of popular lectures on 
astronomy, he so excited the public interest that the necessary funds 



BIOGRAPHY. 465 

were immediately supplied, for erecting an astronomical observatory at 
Harvard College. A remarkable series of lectures on "Ideality in Science/' 
delivered by him in 1879 before the Lowell Institute in Boston, attracted 
the general attention of American thinkers, on account of the thoughtful 
consideration of the vexed question of science and religion. 

Professor Peirce was a transcendentalist in mathematics, as Agassiz 
was in zoology ; and a certain subtile tie of affinity connected these two 
great men, however unlike they were in their special genius. Alike, also, 
they were in their enthusiasm which neither the piercing scepticism of 
Cambridge could wither, nor declining years chill with the frost of age. 
The thing he distrusted was routine and fanatical method, whether new 
or old ; for thought, salient, vital, co-operative thought, in novel or in 
ancient aspects, he had nothing but respect and furtherance. Few men 
could suggest more while saying so little, or stimulate so much while 
communicating next to nothing that was tangible and comprehensible. 
The young man who would learn the true meaning of apprehension as 
distinct from comprehension, should have heard the professor lecture, after 
reciting to him. He was always willing to be esteemed for less than he 
had really accomplished; and he could join most heartily in the praise 
of others who even owed their impulse to him. Modest and magnanimous, 
but not unobservant, his ambition for personal distinction was early and 
easily satisfied ; and he thus rid himself of what is to most men a per- 
turbing, and too often an ignoble, element of discomfort. 

Professor Peirce habitually ascribed to his listener a power of assimi- 
lation which the listener rarely possessed. He assumed his readers could 
follow wherever he led; and this made his lectures hard to follow, his 
books brief, difficult, and comprehensive. When, however, his listeners 
were students who had previously attained some skill as mathematicians 
and who had been trained in his own methods, the resulting work would 
be of the highest order of excellence. He was personally magnetic in his 
presence. His pupils loved and revered him; and to the young man, he 
always lent a helping hand in science. He inspired in them a love of 
truth for its own sake. 

His own faith in Christianity had the simplicity of a child's ; and 
whatever radiance could emanate from a character which combined the 
greatest intellectual attainment with the highest moral worth, that radi- 
ance cast its light upon those who were in his presence. "Every portion 
of the material universe," writes Professor Peirce, "is pervaded by the 
same laws of mechanical action which are incorporated into the very con- 
stitution of the human mind." To him, then, the universe was made 
for the instruction of man. With this belief he approached the study of 
natural phenomena not in the spirit of a critic, but reverently in the 
mood of a sympathizing reader and the lesson he reads is : "There is 
but one God, and science is the knowledge of Him." In his lectures and 
teaching he showed, as he always felt with adoring awe, that the mathe- 
matician enters (as none else can) into the intimate thought of God, sees 
things precisely as they are seen by the Infinite Mind, holds the scales 
and compasses with which the Eternal Wisdom built the earth and meted 
out the heavens. This consciousness had pervaded his whole scientific 
life. It was active in his early youth, as his coevals well remember; it 
gathered strength with his years; and it struck the ever recurring key- 
note in his latest public utterances. 

Benjamin Peirce was a devout, God-fearing man; he was a Christian, 
in the whole aim, tenor, and habit of his life. To know Professor Peirce 
was simply to love him, to admire him, and to revere him. Since he was 
conversant with the phases of scientific infidelity, and by no means un- 
familiar with the historic grounds of scepticism, it can not be regarded 
otherwise than with the profoundest significance, that a mind second to 
none in keen intuition, in aesthetic sensibility, in imaginative fervor, and 
in the capacity of close and cogent reasoning, maintained through life an 



466 FINKEIv'S SOLUTION BOOK. 

unshaken belief and trust in the power, providence, and love of God, as> 
beheld in his works, and as incarnate in our Lord and Savior. In one 
of his lectures on Ideality in Science, he said : " 'Judge the tree by its 
fruit.' Is this magnificent display of ideality a human delusion? Or is 
it a divine record? The heavens and the earth have spoken to declare 
the glory of God. It is not a tale told by an idiot, signifying nothing. It 
is the poem of an infinite imagination, signifying immortality." 

In May, 1880, Professor Peirce began to pass under the shadow of 
the cloud of his last illness. For some weeks there was little serious fear 
that it was a shadow not destined to lift. He was first confined to his 
chamber, on the 25th of June, 3880; and from that time, his slowly failing 
condition was hardly relieved even by any deceptive appearances of im- 
provement. He died on the morning of Wednesday, October 6, 1880. 
Distinguished throughout his life by his freedom from the usual abhor- 
rence of death, which he never permitted himself either to mourn when, 
it came to others, or to dread for himself, he kept this characteristic temper 
to the end, through all the sad changes of his trying illness ; and, two 
days before he ceased to breathe, it struggled into utterance in a few 
faintly-whispered words, which expressed and earnestly inculcated a cheer- 
ful and complete acceptance of the will of God with regard to him. 

The funeral took place on Saturday, October 9, 1880, at Appleton 
Chapel, and was the occasion of an impressive gathering of people of great 
and various mark. The attendance included a very full representation of 
the various faculties and governing boards of the University; a large 
deputation of officers of the United States Coast and Geodetic Survey, 
headed by the superintendent and the chief assistant; delegations of emi- 
nent professors from Yale College and the Johns Hopkins University; 
many members of the class of 1829; and a great number of other friends 
of the deceased. 

The pall-bearers were: President Charles W. Eliot; Ex-President 
Thomas Hill, Pastor of the First Parish Church, Portland, Maine; Capt. 
C. P. Patterson, Superintendent of the United States Coast Survey; Pro- 
fessor J. J. Sylvester, of the Johns Hopkins University ; Hon. J. Ingersoll 
Bowditch ; Professor Simon Newcomb, Superintendent of the American 
Ephemeris and Nautical Almanac; Dr. Oliver Wendell Holmes; Pro- 
fessor Joseph Levering ; and Dr. Merrill Wyman. A beautiful and simple 
service was conducted by the Rev. A. P. Peabody and the Rev. James 
Freeman Clarke. 

In the career of Professor Benjamin Peirce, America has nothing to 
regret, but that it is now closed; while the American people have much 
to learn from his long, useful, and honorable life. By. F. P. Matz. 
From the American Mathematical Monthly. 

BENJAMIN PEIRCE. 

For him the Architect of all 
Unroofed our planet's starlit hall ; 
Through voids unknown to worlds unseen 
His clearer vision rose serene. 

With us on earth he walked by day, 
His midnight path how far away ! 
We knew him not so well who knew 
The patient eyes his soul looked through ; 

For who his untrod realm could share 
Of us that breathe this mortal air, 
Or camp in that celestial tent 
Whose fringes gild our firmament? 

How vast the workroom where he brought 
The viewless implements of thought! 
The wit how subtle, how profound, 
That Nature's tangled webs unwound ; 



BIOGRAPHY. 467 



That through the clouded matrix saw 

The chrystal planes of shaping law, 

Through these the sovereign skill that planned,' 

The Father's care, the Master's hand ! 

To him the wandering stars revealed 
The secrets in their cradle sealed ; 
The far-off, frozen sphere that swings 
Through ether, zoned with lucid rings; 

The orb that rolls in dim eclipse 
Wide wheeling round its long ellipse, 
His name Urania writes with these 
And stamps it on her Pleiades. 

We knew him not? Ah, well we knew 
The manly soul, so brave, so true, 
The cheerful heart that conquered age, 
The childlike silver-bearded sage. 

No more his tireless thought explores 
The azure sea with golden shores ; 
Rest, wearied frame ! the stars shall keep 
A loving watch where thou shalt sleep. 

Farewell ! the spirit needs must rise, 
So long a tenant of the skies, 
Rise to that home all worlds above 
Whose sun is God, whose light is love. 

Oliver Wendell Holmes. 



463 FINKEL'S SOLUTION BOOK. 

BIOGRAPHY. 



JAMES JOSEPH SYLVESTER, LL. D., F. R. S. 

On Monday, March 15, 1897, in London, where, September 3, 1814, he 
was born, died the most extraordinary personage for half a century in the 
mathematical world. 

James Joseph Sylvester was second wrangler at Cambridge in 1837. 
When we recall that Sylvester, Wm. Thomson, Maxwell, Clifford, J. J. 
Thomson were all second wranglers, we involuntarily wonder if any senior 
wrangler except Cayley can be ranked .with them. 

Yet it was characteristic of Sylvester that not to have been first was 
always bitter to him. 

The man who beat him, Wm. N. Griffin, also a Johnion, afterwards 
a modest clergyman, was tremendously impressed by Sylvester, and hon- 
ored him in a treatise on optics where he used Sylvester's first published 
paper, "Analytical development of Fresnel's optical theory of crystals," 
Philosophical Magazine, 1837. 

Sylvester could not be equally generous, and explicitly rated above 
Griffin the fourth wrangler George Green, justly celebrated, who died 
in 1841. 

Sylvester's second paper, "On the motion and rest of fluids/'P/w/o- 
sophical Magazine, 1838 and 1839, also seemed to point to physics. 

In 1838 he succeeded the Rev. Wm. Ritchie as professor of natural 
philosophy in University College, London. 

His unwillingness to submit to the religious tests then enforced at 
Cambridge and to sign the 39 articles not only debarred him from his 
degree and from competing for the Smith's prizes, but, what was far 
worse, deprived him of the Fellowship morally his due. He keenly felt the 
injustice. 

In his celebrated address at the Johns Hopkins University his denun- 
ciation of the narrowness, bigotry and intense selfishness exhibited in these 
compulsory creed tests, made a wonderful burst of oratory. These opin- 
ions were fully shared by De Morgan, his colleague at University Col- 
lege. Copies I possess of the five examination papers set by Sylvester at 
the June examination, session of 1839-40, show him striving as a physicist, 
but it was all a false start. Even his first paper shows he was always 
the Sylvester we knew. To the "Index of Contents" he appends the char- 
acteristic note : "Since writing this index I have made many additions 
more interesting than any of the propositions here cited, which will 
appear toward the conclusion." Ever he is borne along helpless but 
ecstatic in the ungovernable flood of his thought. 

A physical experiment never suggests itself to the great mental experi- 
menter. Cayley once asked for his box of drawing instruments. Sylves- 
ter answered, "I never had one." Something of this irksomeness of the 
outside world, the world of matter, may nave made him accept, in 1841, 
the professorship offered him in the University of _ Virginia. 

On his way to America he visited Rowan Hamilton at Dublin in that 
observatory where the maker of quaternions was as out of place as Syl- 
vester himself would have been. The Virginians so utterly failed to un- 
derstand Sylvester, his character, his aspirations, his powers, that the 
Rev. Dr. Dabney, of Virginia, has seriously assured me that Sylvester' 
was actually deficient in intellect, a sort of semi-idiotic calculating boy. 
For the sake of the contrast, and to show the sort of civilization in which 
this genius had risked himself, two letters from Sylvester's tutors at Cam- 
bridge may here be of interest. 

The great Colenso, Bishop of Natal, previously Fellow and Tutor of 
St. John's College, writes : "Having been informed that my friend and 



BIOGRAPHY. 469 

former pupil, Mr. J. J. Sylvester, is a candidate for the office of professor 
of mathematics, I beg to state my high opinion of his character both as a 
mathematician and a gentleman. 

"On the former point, indeed, his degree of Second Wrangler, at the 
University of Cambridge would be, in itself, a sufficient testimonial. But 
I beg to add that his powers are of a far higher order than even that 
degree would certify." 

Philip Kelland, himself a Senior Wrangler, and then professor of 
mathematics in the University of Edinburgh, writes : "I have been re- 
quested to express my opinion of the qualifications of Mr. J. J. Sylvester, 
as a mathematician. 

"Mr. Sylvester was one of my private pupils in the University of Cam- 
bridge, where he took the degree of Second Wrangler. My opinion of 
Mr. Sylvester then was that in originality of thought and acuteness of 
perception he had never been surpassed, and I predicted for him an emi- 
nent position among the mathematicians of Euiope. My anticipations 
have been verified. Mr. Sylvester's published papers manifest a depth and 
originality which entitles them to the high position they occupy in the 
field of scientific discovery. They prove him to be a man able to grapple 
with the most difficult mathematical questions and are satisfactory evidence 
of the extent of his attainments and the vigor of his mental powers." 

The five papers produced in this year, 1841, before Sylvester's depart- 
ure for Virginia, show that now his keynote is really struck. They 
adumbrate some of his greatest discoveries. 

They are : "On the relation of Sturm's auxiliary functions to the roots 
of an algebraic equation," British Assoc. Rep. (pt. 2), 1841; "Examples 
of the dialytic method of elimination as applied to ternary systems of 
equations," Camb. M. Jour. II., 1841 ; "On the amount and distribution 
of multiplicity in an algebraic equation," Phil. Mag. XyiL, 1841 ; "On 
a new and more general theory of multiple roots," Phil. Mag. XVIIL, 
1841 ; "On a linear method of eliminating between double, treble and other 
systems of algebraic equations," Phil. Mag. XVIIL, 1841 ; "On the dialy- 
tic method of elimination," Phil. Mag. XXL, Irish Acad. Proc. II. 

This was left behind in Ireland, on the way to Virginia. Then sud- 
denly occurs a complete stoppage in this wonderful productivity. Not one 
paper, not one word, is dated from the University of Virginia. Not until 
1844 does the wounded bird begin again feebly to chirp, and indeed it is a 
whole decade before the song pours forth again with mellow vigor that 
wins a waiting world. 

Disheartening was the whole experience ; but the final cause of his 
sudden abandonment of the University of Virginia I gave in an address 
entitled, "Original Research and Creative Authorship the Essence of Uni- 
versity Teaching," printed in Science, N. S., Vol. I., pp. 203-7, February 
22, 1895. 

On the return to England with heavy heart and dampened ardor, he 
takes up for his support the work of an actuary and then begins the study 
of law. In 1847 we find him at 26 Lincoln's Inn Fields, "eating his terms." 
On November 22, 1850, he is called to the bar and practices conveyancing. 

But already in his paper dated August 12, 1850, we meet the significant 
names Boole, Cayley, and harvest is at hand. 

The very words which must now be used to say what had already hap- 
pened and what was now to happen were not then in existence. They 
were afterward made by Sylvester and constitute in themselves a tre- 
mendous contribution. As he himself says: "Names are, of course, all 
important to the progress of thought, and the invention of a really good 
name, of which the want, not previously perceived, is recognized^ when 
supplied, as having ought to be felt, is entitled to rank on a level in im- 
portance, with the discovery of a new scientific theory." 

Elsewhere he says of himself: "Perhaps I may without immodesty 
lay claim to the appellation of the Mathematical Adam, as I believe that 



470 FINKEI/S SOLUTION BOOK. 

I have given more names (passed into general circulation) to the creat- 
ures of the mathematical reason than all the other mathematicians of the 
age combined." 

In one year, 1851, Sylvester created a whole new continent, a new 
world in the universe of mathematics. Demonstration of its creation is 
given by the Glossary of New Terms which he gives in the Philosophical. 
Transactions, Vol. 143, pp. 543-548. 

Says Dr. W. Franz Meyer in his exceedingly valuable Bericht iiber 
die Fortschritte der projectiven Invariantentheorie, the best history of 
the subject (1892) : 

"Als ausseres Zeichen fur den Umfang der vorgeschrittenen Entwicke- 
lung mag die ausgedehnte, grosstenteils von Sylvester selbst herriihrende 
Terminologie dienen, die sich am Ende seiner grossen Abhandlung iiber 
Sturm'sche Functionen (1853) zusammengestellt findet." 

Using then this new language, let us briefly say what had happened 
in the decade when Sylvester's genius was suffering from its Virginia 
wound. The birthday of the giant Theory of Invariants is April 28, 1841, 
the date attached by George Boole to a paper in the Cambridge Mathe- 
matical Journal where he not only proved the invariantive property of dis- 
criminants generally, but also gave a simple principle to form simultaneous 
invariants of a system of two functions. The paper appeared in Novem- 
ber, 1841, and shortly after, in February, 1842, Boole showed that the 
polars of a form lead to a broad class of covariants. Here he extended 
the results of the first article to more than two Forms. Boole's papers 
led Cayley, nearly three years later (1845), to propose to himself the 
problem to determine a priori what functions of the coefficients of an 
equation possess this property of invariance, and he discovered its pos- 
session by other functions besides discriminants, for example the quad- 
rinvariants of binary quantics, and in particular the invariant S of a 
quartic. 

Boole next discovered the other invariant T of a quartic and the ex- 
pression of the discriminant in terms of S and T. Cayley next (1846) 
published a symbolic method of finding invariants. Early in 1851 Boole 
reproduced, with additions, his paper on Linear Transformations; then 
at last began Sylvester. He always mourned what he called "the years 
he lost fighting the world" ; but, after all, it was he who made the 
Theory of Invariants. 

Says Meyer : "sehen wir in dem Cyklus Sylvester'scher Publicationen 
(1851-1854) bereits die Grundzuge einer allegemeinen Theorie erstehen, 
welche die Elemente von den verschiedenartigsten Zweigen der spateren 
Disciplin umfasst." "Sylvester beginnt damit, die Ergebnisse seiner Vor- 
ganger unter einem einzigen Gesichtspunkte zu vereinigen." 

With deepest foresight Sylvester introduced, together with the original 
variables, those dual to them, and created the theory of contravariants and 
intermediate forms. He introduced, with many other processes for pro- 
ducing invariantive forms, the principle of mutual differentiation. 

Hilbert attributes the sudden growth of the theory to these processes 
for producing and handling invariantive creatures. "Die Theorie dieser 
Gebilde erhob sich, von speciellen Aufgaben ausgehend, rasch zu grosser 
Allgemeinheit dank vor Allem dem Umstande, dass es gelang, eine 
Reihe von besonderen der Invariantentheorie eigenthumlichen Prozessen 
zu entdecken, deren Anwendung die Aufstellung und Behandlung invari- 
anter Bildungen betrachtlich erleichterte." 

"Was die Theorie der algebraischen Invarianten anbetrifft so sind die 
ersten Begriinder derselben, Cayley und Sylvester, zugleich auch als die 
Vertreter der naiven Periode anzusehen : an der Aufstellung der einfach- 
sten Invariantenbildungen und an den eleganten Anwendungen auf die 
Auflosung der Gleichungen der ersten 4 Grade hatten sie die unmittelbare 
Freude der ersten Entdeckung." It was Sylvester alone who created the 



BIOGRAPHY. 471 

theory of canonic forms and proceeded to apply it with astonishing power. 
What marvelous mass of brand new being he now brought forth ! 

Moreover he trumpeted abroad the eruption. He called for communi- 
cations to himself in English, French, Italian, Latin or German, so only 
the "Latin character" were used. 

From 1851 to 1854 he produces forty-six different memoirs. Then 
comes a dead silence of a whole year, broken in 1856 by a feeble chirp 
called "A Trifle on Projectiles." 

What has happened? Some more "fighting the world." Sylvester 
declared himself a candidate for the vacant professorship of geometry in 
Gresham College, delivered a probationary lecture on the 4th of Decem- 
ber, 1854, and was ignominiously "turned down." Let us save a couple of 
sentences from this lecture : 

"He who would know what geometry is must venture boldly into its 
depths and learn to think and feel as a geometer. I believe that it is 
impossible to do this, to study geometry as it admits of being studied, and 
I am conscious it can be taught, without finding the reasoning invigorated, 
the invention quickened, the sentiment of the orderly and beautiful awak- 
ened and enhanced, and reverence for truth, the foundation of all integrity 
of character, converted into a fixed principle of the mental and moral 
constitution, according to the old and expressive adage 'abeunt studia in 
mores/ " 

But this silent year concealed still another stunning blow of precisely 
the same sort, as bears witness the f611owing letter from Lord Brougham 
to The Lord Panmure: 

" BROUGHAM, 

PRIVATE. 28 Aug. 1855. 

MY DEAR P. 

My learned excellent friend and brother mathematician Mr. Sylvester is again a 
candidate for the professorship at Woolwich on the death of Mr. O'Brian who carried it 
against him last year. 

I entreat once more your favorable consideration of this eminent man who has al- 
ready to thank you for your great kindness. 

Yours sincerely, 

H. BROUGHAM. 

On this third trial, backed by such an array of credentials as no man 
ever presented before, he barely scraped through, was appointed professor 
of mathematics at the Royal Military Academy, and served at Woolwich 
exactly 14 years, 10 months, and 15 days. 

A single sentence of his will best express his greatest achievement 
there and his manner of exit thence : 

"If Her most Gracious Majesty should ever be moved to recognize 
the palmary exploit of the writer of this note in the field of English science 
as having been the one successfully to resolve a question and conquer an 
algebraical difficulty which had exercised in vain for two centuries past, 
since the time of Newton, the highest mathematical intellects in Europe 
(Euler, Lagrange, Maclaurin, Waring among the number), by conferring 
upon him some honorary distinction in commemoration of the deed, he 
will crave the privilege of being allowed to enter the royal presence, not 
covered, like De Courcy, but barefooted, with rope around his waist, and 
a goose-quill behind his ear, in token of repentant humility, and as an 
emblem of convicted simplicity in haying once supposed that on such 
kind of success he could found any additional title to receive fair and just 
consideration at the hands of Her Majesty's Government when quitting 
his appointment as public professor at Woolwich under the coercive opera- 
tion of a non-Parliamentary retrospective and utterly unprecedented War 
Office enactment." Athenaeum Club, January 31, 1871. Of course this 
means a row of barren years, 1870, 1871, 1872, 1873. 

The fortunate accident of a visit paid Sylvester in the autumn of 1873 
by Pafnuti Lvovich Chebyshev, of the University of St. Petersburg, re- 
awakened our genius to produce in a single burst of enthusiasm a new- 
branch of science. 



472 FINKEVS SOLUTION BOOK. 

On Friday evening, January 23, 1874, Sylvester delivered at the Royal 
Institution a lecture entitled "On Recent Discoveries in Mechanical Con- 
version of Motion," whose ideas, carried on by two of his hearers, H. 
Hart and A. B. Kempe, have made themselves a permanent place even 
in the elements of geometry and kinematics. A synopsis of this lecture 
was published, but so curtailed and twisted into the third person that the 
life and flavor are quite gone from it. I possess the unique manuscript 
of this epoch-making lecture as actually delivered. A few sentences will 
show how characteristic and inimitable was the original form: 

"The air of Russia seems no less favorable to mathematical acumen 
than to a genius for fable and song. Lobacheffsky, the first to mitigate 
the severity of the Euclidean code and to beat down the bars of a supposed 
adamantine necessity, was born (a Russian of Russians), in the govern- 
ment of Nijni Novgorod; Tchebicheff [Chebyshev], the prince and con- 
queror of prime numbers, able to cope with their refractory character and 
to confine the stream of their erratic flow, their progression, within alge- 
braic limits, in the adjacent circumscription of Moscow; and our own 
Cayley was cradled amidst the snows of St. Petersburg." [Sylvester him- 
self contracted Chebyshev's limits for the distribution of primes.] "I 
think I may fairly affirm that a simple direct solution of the problem of 
the duplication of the cube by mechanical means was never accomplished 
down to this day. .1 will not say but that, by a merciful interpretation of 
his oracle, Apollo may have put up with the solution which the ancient 
geometers obtained by means of drawing two parabolic curves ; but of 
this I feel assured that had I been then alive, and could have shown my 
solution, which I am about to exhibit to you, Apollo would have leaped 
for joy and danced (like David before the ark), with my triple cell in 
hand, in place of his lyre, before his own duplicated altar." 

That in the very next year Sylvester was taking a more active part 
than has hitherto been known in the organization of the incipient Johns 
Hopkins University is seen from the following letter to him in London 
from the great Joseph Henry : 

SMITHSONIAN INSTITUTION, 

August 25, 1875. 
MY DEAR SIR: 

Your letter of the 13th inst. has just been received and in reply I have to say that I 
have written to President Gilman of the Hopkins Universit}' giving my views as to what 
it ought to be and have stated that if properly managed it may do more for the advance 
of literature and science in this country than any other institution ever established ; it 
is entirely independent of public favor and may lead instead of following popular 
opinion. 

I have advised that liberal salaries be paid to the occupants of the principal chairs 
and that to fill them the best men in the world who can be obtained should be secured. 

I have mentioned your name prominently as one of the very first mathematicians 
of the day ; what the result will be, however, I can not say. 

The Trustees are all citizens of Baltimore and among them I have some personal 
friends ; the President, Mr. Gilman, and one of them, came to Washington a few weeks 
ago to get from me any suggestions that I might have to offer. 

It is to be regretted that in this country the Trustees, who control the management 
of bequests of this character, think it important to produce a palpable manifestation of 
the institution to be established by spending a large amount of the bequest in archi- 
tectural displays. Against this custotn I have protested and have asserted that if the 
E roper men and necessary implements of instruction are provided, the teaching may 
e done in log cabins. 

It would give me great pleasure to have you again as my guest, and I will do what I 
can to secure your election. Very truly your friend, 

JOSEPH HENRY. 
We know the result. 

Sylvester was offered the place; demanded a higher salary; won; 
came. 

I was his first pupil, his first class, and he always insisted that it was 
I who brought him back to the Theory of Invariative Forms. In a letter 
to me of September 24, 1882, he writes : "Nor can I ever be oblivious of 
the advantage which I derived from your well-grounded persistence in 
inducing me to lecture on the Modern Algebra, which had the effect of 
bringing my mind back to this subject, from which it had for some time 



BIOGRAPHY. 47a 

previously been withdrawn, and in which I have been laboring, with a 
success which has considerably exceeded my anticipations, ever since." 

He made this same statement at greater length in his celebrated ad- 
dress at the Johns Hopkins on February 22, 1877: "At this moment I 
happen to be engaged in a research of fascinating interest to myself, and 
which, if the day only responds to the promise of its dawn, will meet, I 
believe, a sympathetic response from the professors of our divine alge- 
braical art wherever scattered through the world. 

"There are things called Algebraical Forms ; Professor Cayley calls 
them Quantics. These are not, properly speaking, Geometrical Forms, al- 
though capable, to some extent, of being embodied in them, but rather 
schemes of processes, or of operations for forming, for calling into exist- 
ence, as it were, algebraic quantities. 

"To every such Quantic is associated an infinite variety oi other forms 
that may be regarded as engendered from and floating, like an atmosphere, 
around it ; but infinite in number as are these derived existences, these 
emanations from the parent form, it is found that they admit of being 
obtained by composition, by mixture, so to say, of a certain limited num- 
ber of fundamental forms, standard rays, as they might be termed, in the 
Algebraic Spectrum of the Quantic to which they belong; and, as it is a 
leading pursuit of the physicists of the present day to ascertain the fixed 
lines in the spectrum of every chemical substance, so it is the aim and 
object qf a great school of mathematicians to make out the fundamental 
derived forms, the Covariants and Invariants, as they are called, of these 
Quantics. 

"This is the kind of investigation in which I have, for the last month 
or two, been immersed, and which I entertain great hopes of bringing 
to a successful issue. 

"Why do I mention it here? It is to illustrate my opinion as to the 
invaluable aid of teaching to the teacher, in throwing him back upon his 
own thoughts and leading him to evolve new results from ideas that 
would have otherwise remained passive or dormant in his mind. 

"But for the persistence of a student of this university in urging upon 
me his desire to study with me the modern algebra I should nev k er have 
been led into this investigation ; and the new facts and principles which 
I have discovered in regard to it (important facts, I believe) would, so 
far as I am concerned, have remained still hidden in the womb of time. 
In vain I represented to this inquisitive student that he would do better 
to take up some other subject lying less off the beaten track of study, 
such as the higher parts of the Calculus or Elliptic Functions, or the theory 
of Substitutions, or I wot not what besides. He stuck with perfect re- 
spectfulness, but with invincible pertinacity, to his point. He would have 
the New Algebra (Heaven knows where he had heard about it, for it is 
almost unknown on this continent), that or nothing. I was obliged to 
yield, and what was the consequence? In trying to throw light upon an 
obscure explanation in our text-book my brain took fire ; I plunged with 
requickened zeal into a subject which I had for years^ abandoned, and 
found food for thoughts which have engaged my attention for a consider- 
able time past, and will probably occupy all my powers of contemplation 
advantageously for several months to come." 

Another specific instance of the same thing he mentions in his paper, 
"Proof of the Hitherto Undemonstrated Fundamental Theorem of Inyari- 
ants," dated November 13, 1877: 

"I am about to demonstrate a theorem which has been waiting proof 
for the last quarter of a century and upwards. It is the more necessary 
that this should be done, because the theorem has been supposed to lead 
to false conclusions, and its correctness has consequently been impugned. 
Thus in Professor Faa de Bruno's valuable Theorie des formes binaires, 
Turin, 1876, at the foot of page 150 occurs the following passage: ''Cela 
suppose essentiellement que les equations de condition soient toutes inde'- 



474 FlNKEIv'S SOLUTION BOOK. 

pendantes entr'elles, ce qui n'est pas toujours le cas, ainsi qu'il resulte 
des recherches du Professor Gordan sur les nombres des covariants des 
formes quintique et sextique." 

The reader is cautioned against supposing that the consequence alleged 
above does result from Gordan's researches, which are indubitably correct. 
This supposed consequence must have arisen from a misapprehension, on 
the part of M. de Bruno, of the nature of Professor Cayley's rectification 
of the error of reasoning contained in his second memoir on Quantics, 
which had led to results discordant with Gordan's. Thus error breeds 
error, unless and until the pernicious brood is stamped out for good and 
all under the iron heel of rigid demonstration. In the early part of this 
year Mr. Halsted, a fellow of Johns Hopkins University, called my atten- 
tion to this passage in M. de Bruno's book; and all I could say in reply 
was that 'the extrinsic evidence in support of the independence of the equa- 
tions which had been impugned rendered it in my mind as certain as any 
fact in nature could be, but that to reduce it to an exact demonstration 
transcended, I thought, the powers of the human understanding.' " 

In 1883 Sylvester was made Savilian professor of geometry at Oxford, 
the first Cambridge man so honored since the appointment of W.allis in 
1649. 

To greet the new environment, he created a new subject for his re- 
searches Reciprocants, which has inspired, among others, J. Hammond, 
of Oxford; McMahon, of Woolwich; A. R. Forsyth, of Cambridge; 
Leudesdorf, Elliott and Halphen. 

Sylvester never solved exercise problems such as are proposed in the 
Educational Times, though he made them all his life long down to his 
latest years. For example, unsolved problems by him will be found even 
in Vol. LXIL and Vol. LXIII. of the Educational Times reprints (1395). 
If at the time of meeting his own problem he met also a neat solution 
he would communicate them together, but he never solved any. In the 
meagre notices that have been given of Sylvester the strangest errors 
abound. Thus C. S. Pierce, in the Post, March 16th, speaks of his 
accepting, "with much diffidence," a word whose meaning he never knew; 
and gives 1862 as the date of his retirement from Woolwich, which is 
eight years wrong, as this forced retirement was July 31, 1870, after his 
55th birthday. Cajori, in his inadequate account (History of Mathematics, 

&326), puts the studying of law before the professorship at University 
^ allege and the professorship at the University of Virginia, both of which 
it followed. Effect must follow cause. And strange, that of the few 
things he ascribes to Sylvester, he should have hit upon something not 
his, "the discovery of the partial differential equations satisfied by the 
invariants and covariants of binary quantics." But Sylvester has ex- 
plicitly said in Section VI. of his "Calculus of Forms" : "I alluded to 
the partial differential equations by which every invariant may be de- 
fined. M. Aronhold, as I collect from private information, was the first 
to think of the application of this method to the subject; but it was Mr. 
Cayley who communicated to me the equations which define the invariants 
of functions of two variables." 

Surely he needs nothing but his very own, this marvellous man who 
gave so lavishly to every one devoted to mathematics, or, indeed, to the 
highest advance of human thought in any form. By George Bruce Hal- 
sted. From the American Mathematical Monthly. 



BIOGRAPHY. 475 



BIOGRAPHY. 



ARTHUR CAYLEY. 

Arthur Cayley was born at Richmond in Surrey, England, August 
the 16th, 1821. His father, Henry Cayley, was descended from the Cay- 
leys of Brompton, in Yorkshire, but was at the time a merchant of St. 
Petersburg where he had married a Russian lady. In 1829 his parents 
took up their permanent residence at Blackheath in England ; and Arthur 
was there educated at a private school for four years. At the age of 14 
he was sent to King's College School, London; and the master of that 
school having observed the promise of a mathematical genius advised 
the father to educate his son not for his own business, but to enter the 
University of Cambridge. 

In 1838 Arthur Cayley entered Trinity College, Cambridge, at the 
rather early age of 17. Throughout his undergraduate course he was 
first at his college examinations by an enormous interval, and he finished 
his undergraduate career in 1842 by carrying off the two highest honors, 
namely, the first place, or Senior Wrangler, in the Mathematical Tripos, 
and the first prize in the competition for the Smith Prizes. Immediately 
elected a Fellow of his College, he continued to reside at Cambridge for 
several years, during which time he lectured on mathematics, and also 
contributed papers to the Cambridge Mathematical Journal. His first 
contribution to that Journal was made, when he was an undergraduate, 
in 1841. 

At that time it was necessary for a Fellow to take Holy Orders, or 
else resign the fellowship at the end of seven years. Mr. Cayley chose the 
latter alternative, and became by profession a conveyance in Lincoln's Inn, 
London. He followed that profession for 14 years with conspicuous ability 
and success, and at the same time made many of his most important con- 
tributions to mathematical science. 

About 1861 the Lucasian professorship of mathematics at Cambridge 
the chair made illustrious by Sir Isaac Newton fell vacant; it was 
filled by G. G. Stokes, already eminent for his work in mathematical 
physics, and Senior Wrangler the year before Cayley. However, it was 
felt desirable to secure Cayley also, and for this purpose the -Sadlerian 
professorship of mathematics was created, which resulted in Cayley marry- 
ing and settling down at Cambridge, in 1863. 

The duties of the Sadlerian professor were defined as follows : "to 
explain and teach the principles of pure mathematics, and to apply himself 
to the advancement of the science." In carrying out the former part of 
the duties Professor Cayley did not give the same course of lectures year 
after year, but each year took for his subject that of the memoir on 
which he was engaged. As a consequence his students were few, for 
advanced work of that kind did not pay in the great mathematical exam- 
ination. How well he carried out the second part of the duties may be 
inferred from the fact that the Royal Society Catalogue of Scientific 
Papers enumerates 430 memoirs contributed by him between the years 
1863 and 1883, making a total up to the latter date of 724. As he con- 
tinued active to the last, it is probable that the grand total of his papers 
does not fall short of 1000. Some of his most celebrated contributions 
are: Chapters in the Analytical Geometry of (n) Dimensions, On the 
theory of Determinants, On the theory of linear transformations, Ten 
Memoirs on Quantics, Memoir on the theory of Matrices, Memoirs on 
Skew Surfaces, otherwise Scrolls, On the Motion of Rotation of a Solid 
Body, On the triple tangent planes of surfaces of the third order. Several 
of his achievements are elegantly referred to in a poem written by his 



476 FINKEI/S SOLUTION BOOK. 

colleague Clerk Maxwell in 1874, and addressed to the Committee of 
subscribers who had charge of the Cayley Portrait Fund : 

O wretched race of men, to space confined ! 

What honor can ye pay to him whose mind 

To that which lies beyond hath penetrated? 

The symbols he hath formed shall sound his praise, 

And lead him on through unimagiued ways 

To conquests new, in worlds not yet created. 

First, ye Determinants, in order row 

And massive column ranged, before him go, 

To form a phalanx for his safe protection, 

Ye powers of the nth root of 1 ! 

Around his head in endless cycles run 

As unembodied spirits of direction. 

And you, ye undevelopable scrolls ! 

Above the host wave your emblazoned rolls, 

Ruled for the record of his bright inventions. 

Ye cubic surfaces ! by threes and nines 

Draw round his camp your seven and twenty lines 

The seal of Solomon in three dimensions. 

March on, symbolic host ! with step sublime, 
Up to the naming bounds of Space and Time ! 
There pause, until by Dickenson depicted, 
In two dimensions, we the form may trace 
Of him whose soul, top large for vulgar space, 
In n dimensions flourished unrestricted. 

The portrait was presented to Trinity College, and now adorns their 
Hall. He is represented as seated at a desk, with quill in hand, and think- 
ing out intently some mathematical idea. 

But mathematical science was advanced by Professor Cayley in yet 
another way. By his immense learning, his impartial judgment, and his 
friendly sympathy with other workers, he was eminently qualified to act 
as a referee on mathematical papers contributed to the various societies. 
Of this kind of work he did a large amount, and of his kindliness to young 
investigators I can speak from personal experience. Several papers which 
I read before the Royal Society of Edinburgh were referred to him, and 
he recommended their publication. Some time after I attended a meeting 
of the Mathematical Society of London, but the friend who would have 
introduced me could not be present. Professor Cayley was present, and 
on finding out who I was, gave me a cordial handshake, and referred in 
the kindest terms to the papers he had read. He was a cosmopolitan 
spirit, delighting only in the truth, and friendly to all seekers after the 
truth. 

Among Cayley's papers there are several on a "Question in the 
Theory of Probabilities." The question was propounded by Boole, and he 
applied to its solution the general method of "The Laws of Thought." It 
was afterwards discussed by Wilbraham, Cayley and others in the Philo- 
sophical Magazine. My attention was drawn to the question when writ- 
ing the Principles of the Algebra of Logic, and I ventured to contribute 
my idea of the question to the Educational Times. On mentioning the 
matter to Professor Kelland, he intimated pretty plainly that the discus- 
sion had been closed by Professor Cayley, and that it was temerity on my 
part to write anything on the subject. But the great mathematician did 
not think so; he wrote me a letter discussing the question and my, par- 
ticular way of viewing it, as well as the fundamental ideas in which I 
differed from Boole. 

In 1882 he received a flattering invitation from the trustees of the 
Johns Hopkins University to deliver a course of lectures on some subject 
in advanced mathematics. He chose as his subject the Elliptic and 
Abelian functions ; and the impression which his presence created has 
been well described by Dr. Matz in his brief notice in the January number 
of the MONTHLY. 



BIOGRAPHY. 477 

Next year he was president of the British Association at the South- 
port meeting. In his address he spoke of the foundations of mathematics, 
reviewed the more important theories, traced the connection of pure with 
applied mathematics, and gave an outline of the vast extent of Modern 
Mathematics. 

He regarded the complex number a -f- bi as the fundamental quantity 
of mathematical analysis, and considered that with such a basis, algebra 
was a complete and bounded science, in which no further imaginary sym- 
bols could spring up. It is the more remarkable that he held such a 
view, when we consider that early in his career he made a notable con- 
tribution to space analysis. Starting from Rodrigues' formulae for the 
rotation of a solid body, he arrived at the quaternion formula, and was 
anticipated by Hamilton only by a few months. But Cayley took a 
Cartesian view of analysis to the last, as is evident from the chapter 
which he contributed to Tait's Treatise on Quaternions. His aim there is 
to give an analytical theory of quaternions. Hamilton's aim on the other 
hand was to give a quaternionic theory of analysis. The difference is 
brought out still more strikingly in a paper printed in the last number of 
the Proceedings of the Royal Society of Edinburgh. 

In 1889 the Cambridge University Press commenced the re-publication 
of his mathematical papers in a collected form. It was calculated that 
they would occupy 10 quarto volumes ; 12 volumes have already appeared ; 
and it is believed that 13 volumes will be required. No mathematician has 
ever had his works printed in a more handsome manner. In addition he 
is the author of a separate work on Elliptic Functions. 

Space fails to enumerate the honors which he received from Univer- 
sities and Scientific Academies both of the Old and of the New World. 
But we may mention specially, that from the Royal Society he received 
a Royal Medal and a Copley Medal; from the Mathematical Society of 
London the first DeMorgan Medal; and at the instance of the President 
and Members of the French Academy he was made an Officer of the Legion 
of Honour. 

On the 26th of January he died at Cambridge. His body was laid to 
rest in Mill Road Cemetery in the presence of official representatives from 
foreign countries and many of the most illustrious philosophers of Eng- 
land. His spirit still speaks to us from his works, and will continue to 
speak to many succeeding generations. By Dr. Alexander Macfarlane. 
From the American Mathematical Monthly, Vol. II., No. 4. In the same 
number is also* an interesting biography of Cayley, by Dr. George Bruce 
Halsted. 



478 



FINKEL'S SOLUTION BOOK. 



TABLE I. Functions of n and e. 



n = 3-1415926 
7r 2 = 9-8696044 
7T 3 =31-0062761 
*/*== 1-7724539 
log 10 7T= 1-4971499 
log e ?r= 0-6679358 


jr-i= -3183099 
7T- 2 = -1013212 
7T- 8 = -0322515 
200 s7 -7-^=63* -6619772 
180 -7-7r=57-2957795 
=206264 /x -8 


e =2-71828183 
*2 =7-38905611 
^-1=0-3678794 
^-2=0-1353353 
Iog 10 =0-43429448 
log, 10 =2-30258509 



TABLE II. 



TABLE III. 



No. 


Square root. 


Cube root. 


2 


1-4142136 


1-2599210 


3 


17320508 


1-4422496 


4 


2-0000000 


1-5874011 


5 


2-2360680 


1-7099759 


6 


2-4494897 


1-8171206 


7 


2-6457513 


1-9129312 


8 


2-8284271 


2-0000000 


9 


3-0000000 


2-0800837 


10 


3-1622777 


2-1544347 


11 


3-3166248 


2-2239801 


12 


3-4641016 


2-2894286 


13 


3-6055513 


2-3513347 


14 


3-7416574 


2-4101422 


15 


3-8729833 


2-4662121 


16 


4-0000000 


2-5198421 


17 


4-1231056 


2-5712816 


18 


4-2426407 


2-6207414 


19 


4-3588989 


2-6684016 


20 


4-4721360 


2-7144177 


21 


4-5825757 


2-7589243 


22 


4-6904158 


2-8020393 


23 


4-7958315 


2-8438670 


24 


4-8989795 


2-8844991 


25 


5-0000000 


2-9240177 


26 


5-0990195 


2-9624960 


27 


5-1961524 


3-0000000 


28 


5-2915026 


3-0365889 


29 


5-3851648 


3-0723168 


30 


5-4772256 


3-1072325 



N. 


logio N. 


log* N. 


2 


3010300 


69314718 


3 


4771213 


1-09861229 


5 


6989700 


1-60943791 


7 


8450980 


1-94591015 


11 


1-0413927 


2-39789527 


13 


1-1139434 


2-56494936 


17 


1-2304489 


2-83321334 


19 


1-2787536 


2-94443898 


23 


1-3617278 


3-13549422 


29 


1-4623980 


3-36729583 


31 


1-4913617 


3-43398720 


37 


1-5682017 


3-61091791 


41 


1-6127839 


3-71357207 


43 


1-6334685 


3-76120012 


47 


1-6720979 


3-85014760 


53 


1-7242759 


3-97029191 


59 


1-7708520 


4-07753744 


61 


1-7853298 


4-11087386 


67 


1-8260748 


4-20469262 


71 


1-8512583 


4-26267988 


73 


1-8633229 


4-29045944 


79 


1-8976271 


4-36944785 


83 


1-9190781 


4-41884061 


89 


1-9493900 


4-48863637 


97 


1-9867717 


4-57471098 


101 


2-0043214 


4-61512052 


103 


2-0128372 


4-63472899 


107 


2-0293838 


4-67282883 


109 


2-0374265 


4-69134788 






.* 



TABLES OP LOGARITHMS. 



479 



TABLE IV. The Natural Logarithms (each increased by 10) 
of Numbers between 0.00 and 0.99. 



N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


0.0 




5.395 


6.088 


6.493 


6.781 


7.004 


7.187 


7.341 


7.474 


7.592 


0.1 


7.697 


7.793 


7.880 


7.960 


8.034 


8.103 


8.167 


8.228 


8.285 


8.339 


0.2 


8.391 


8.439 


8.846 


8.530 


8.573 


8.614 


8.653 


8.691 


8.727 


8.762 


0.3 


8.796 


8.829 


8.861 


8.891 


8.921 


8.950 


8.978 


9.006 


9.032 


9.058 


0.4 


9.084 


9.108 


9.132 


9.156 


9.179 


9.201 


9.223 


9.245 


9.266 


9.287 


0.5 


9.307 


9.327 


9.346 


9.365 


9.384 


9.402 


9.420 


9.438 


9.455 


9.472 


0.6 


9.489 


9.506 


9.522 


9.538 


9.554 


9.569 


9.584 


9.600 


9.614 


9.629 


0.7 


9.643 


9.658 


9.671 


9.685 


9.699 


9.712 


9.726 


9.739 


9.752 


9.764 


0.8 


9.777 


9.789 


9.802 


9.814 


9.826 


9.837 


9.849 


9.861 


9.872 


9.883 


0.9 


9.895 


9.906 


9.917 


9.927 


9.938 


9.949 


9.959 


9.970 


9.980 


9.990 



TABLE V. The Natural Logarithms of Numbers between 
1.0 and 9.9. 



N. 





1 


2 


3 


4 


5 


6 


7 


8 


9 


1 


0.000 


0.095 


0.182 


0.262 


0.336 


0.405 


0.470 


0.531 


0.588 


0.642 


2 


0.693 


0.742 


0.788 


0.833 


0.875 


0.916 


0.956 


0.993 


1.030 


1.065 


3 


1.099 


1.131 


1.163 


1.194 


1.224. 


1.253 


1.281 


1.308 


1.335 


1.361 


4 


1.386 


1.411 


1.435 


1.459 


1.482 


1.504 


1.526 


2.548 


1,569 


1.589 


5 


1.609 


1.629 


1.649 


1.668 


1.686 


1.705 


1.723 


1.740 


1.758 


1.775 


6 


1.792 


1.808 


1.825 


1.841 


1.856 


1.872 


1.887 


1.902 


1.917 


1.932 


7 


1.946 


1.960 


1.974 


1.988 


2.001 


2.015 


2.028 


2.041 


2.054 


2.067 


8 


2.079 


2.092 


2.104 


2.116 


2.128 


2.140 


2.152 


2.163 


2.175 


2.186 


9 


2.197 


2.208 


2.219 


2.230 


2.241 


2.251 


2.262 


2.272 


2.282 


2.293 



480 



FINKEL'S SOLUTION BOOK. 



TABLE VI. The Values in Circular Measure of Angles which 
are given in Degrees and Minutes. 



V 


0.0003 


20' 


0.0058 


7 


0.1222 


80 


1.3963 


2' 


0.0006 


30' 


0.0087 


8 


0.1396 


90 


1.5708 


3' 


0.0009 


40' 


0.0116 


9 


0.1571 


100 


1.7453 


4' 


0.0012 


50' 


0.0145 


10 


0.1745 


110 


1.9199 


5' 


0.0015 


60'orl 


0.0175 


20 


0.3491 


120 


2.0944 


6' 


0.0017 


2 


0.0349 


30 


0.5236 


130 


2.2689 


7' 


0.0020 


3 


0.0524 


40 


0.6981 


140 


2.4435 


& 


0.0023 


4 


0.0698 


50 


0.8727 


150 


2.6180 


9' 


0.0026 


5 


0.0873 


60 


1.0472 


160 


2.7925 


10' 


0.0029 


6 


0.1047 


70 


1.2217 


170 


2.9671 



TABLE VII. Equivalents of Radians in Degrees, Minutes, and 
Seconds of Arc. 



Radians. 


Equivalents. 


Radians. 


Equivalents. 


0.0001 


0' 20".6 


0.0600 


3 26' 15".9 


0.0002 


0' 41".3 


0.0700 


4 0' 38".5 


0.0003 


T 01".9 


0.0800 


4 35' 01".2 


0.0004 


1' 22".5 


0.0900 


5 9' 23".8 


0.0005 


1' 43".l 


0.1000 


5 43' 46".5 


0.0006 


2' 03".8 


0.2000 


11 27' 33".0 


0.0007 


2' 24".4 


0.3000 


17 11' 19".4 


0.0008 


2' 45".0 


0.4000 


22 55' 05".9 


0.0009 


3' 05".6 


0.5000 


28 38' 52".4 


0.0010 


3' 26".3 


0.6000 


34 22' 38".9 


0.0020 


6' 52".5 


0.7000 


40 6' 25".4 


0.0030 


10' 18".8 


0.8000 


45 50' 11".8 


0.0040 


13' 45'M 


0.9000 


51 33' 58".3 


0.0050 


17' 11".3 


1.0000 


57 17' 44".8 


0.0060 


20' 37".6 


2.0000 


114 35' 29".6 


0.0070 


24' 03".9 


3.0000 


171 53 V 14".4 


0.0080 


27' 30".l 


4.0000 


229 10' 59".2 


0.0090 


30' 56".4 


5.0000 


286 28' 44".0 


0.0100 


34' 22".6 


6.0000 


343 46' 28".8 


0.0200 


1 8' 45".3 


7.0000 


401 4' 13".6 


0.0300 


1 43' 07".9 


8.0000 


458 21' 58".4 


0.0400 


2 17' 30".6 


9.0000 


515 39' 43".3 


0.0500 


2 51' 53V2 


10.0000 


572 57' 28".l 



EXAMPLE. 



481 



To USE TABLE VII. For example, express 1.3245 radians in 
degrees, minutes, and seconds. 



1. 1.0000 radian == 57 

2. .3000 " =17 

3. .0200 " = 1 

4. .0040 " = 

5. .0005 " 



17' 
11' 

8' 
13' 

V 



44".3 
19".4 
45".3 
45".l 
43".l 



Adding, 1.3245 radians = 77 53' 17".2 



14 DAY USE 

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LD 21A-50m-ll,'62 
(D3279slO)476B 



General Library 

University of California 

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YC 22416